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Full text of "The elements of Euclid, viz. the first six books, together with the eleventh and twelfth. The errors by which Theon, or others, have long ago vitiated these books are corrected, and some of Euclid's demonstrations are restored. Also, the book of Euclid's Data, in like manner corrected. By Robert Simson. 13th ed., carefully rev. and improved, to which is added a treatise on the construction of the trigonometrical canon; and a concise account of logarithms"

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THE 

ELEMENTS 



OF 



EUCLID; 

VIZ, 

THE FIRST SIX BOOKS, 

TOGETHER WITH THE 

ELEVENTH AND TWELFTH. 

The ERRORS by which Theon, or others, have long ago 
vitiated these Books, are Corrected, 

And some of EUCLID'S DemoxNstrations are Restored. 

ALSO 

THE BOOK OF 

EU GLIDES DATA. 

In like Manner Corrected. 



% ROBERT SIMSON, M. D. 

Emeritus Professor of Mathematics in the University of Glasgow. 



The Thiiltiinth Edition, carefully revised and improyed, 
» 

To which is added, 

A TREATISE on the CONSTRUCTION of thk TRIGONOMETRICAL 
CANON i AND A CONCISE ACCOUNT of LOGARITHMS. 

LONDON: 

MINTED FOB r. WISGRAVE; J. JOHKSONJ W. J. AKD J. RICHARDSON; 
F. C. AND J. BIVIXGTON; J. walker; J. SCATCHERD AND C. J, 
tETTERMAN ; C. WILKIE AND J. ROBINSON; VERNOR, HOOD, Alfl> 
SHARPS ; LONGMAN, HURST, REES, AND OBME J T. CADIH AJTO W, 
DATIES; E. PHILUPS; AND J. MAWMAK. 

1806. 



90«184> 



WniGHi", Printn; 
Vo. W, St. John » S<|»ai*, Clertf»W«U. 



^ % 



TO 

THE KING, 

THIS EDITION 

«r TKX 

PRINCIPAL BOOKS 

Ot TU 

ELEMENTS OF EUCLID, 

AX9 or TSX 

BOOK OF HIS DATA, 

IS MOST BVMBLT DSDICATSD, 
BT 

HIS MAJESTY'S 

MOST DUTIFUL, 

AND MOST DBVOTBD 

SUBJECT AND SERVANT, 

ROBERT SIMSON. 



CONTENTS. 

PREFACE, 

The ELEMENTS, * Page i to 2^ 

NOTES, Critical and Geometrical, -- 287. 

The DATA, 355 

Plane TRIGONOMETRY, 473 

CONSTRUCTIONSof the TRIGONOMETRICAL 
CANON, r 48s 

Of LOGARITHMS, 49^ 

Spherical TRIGONOMETRY, 504 



PREFACE. 



Jl HE opinions of the moderns concerning the author of the 
Elements of Geometry, which go under Euclid's name, are 
very different and contrar}' to one another. Peter Ramus as- 
cribes the Propositions, as well as their Demonstrations, to 
Theon ; others think the Propositions to be Euclid's, but that 
the Demonstrations are Theon's ; and others maintain, that 
all the Propositions and their Demonstrations are Euclid's 
own. John Buteo and Sir Henry Savile are the authors of 
greatest note who assert this last ; and the greater part of geo- 
meters have ever since been of this opinion, as they thought it 
the most probable. Sir Henry Savile, after the several argu- 
ments he brings to prove it, makes this conclusion (Page 13. 
Praeleft.) '* That, excepting a very few interpolations, expli- 
" cations, and additions, Theon altered nothing in Euclid." 
But, by often considering and comparing together the Defi- 
nitions and Demonstrations as they are in the Greek editions 
we now have, I found that Theon, or whoever was the editor 
of the present Greek text, by adding some things, suppressing 
others, and mixing his own with Euclid's Demonstrations, 
had changed more things to the worse than is commonly sup- 
posed, an<l those not of small moment, especially in the fifth 
and eleventh Books of the Elements, which this editor has 
greatly vitiated ; for instance, by substituting a shorter, but 
insufficient Demonstration of the iSthProp. ofthe 5th Book, 
in place of the legitinwite one which Euclid had given ; and by 
taking out of this Book, besides other things, the good defi- 
nition which Eudoxus or Euclid had given of compound ratio, 
and giving an absurd one in place of it in the 5th Defini- 
tion of the 6th Book, which neither Euclid, Archimedes, 
Apellotiius, nor any geometer before Theon's time, ever 
made usp of, and of which there is not to be found the least 
appearance in any of their writings j and, as this Definition 
did much embarrass beginners, and is quite useless, it is now 
thrown out of the Elements, and another, which, without 
<ioubt, Euclid had given, is put in its proper place among the 

Definitioo^ 



VI PREFACE. 

Definitions of the 5th Book, by which the doctrine of com- 
pound ratios is rendered plain and easy. Besides among the 
Definitions of the i ith Book, there is this, which is the tenth, 
viz. ** Equal and similar solid figures are those which arc 
** contained by similar planes of the same number and mag- 
" nitude." Now this Proposition is a Theorem, not a Defi- 
nition ; because the equality of figures of any kind must be 
demonstrated, and not assumed ; and therefore, though this 
were a triie Proposition, it ought to have been demonstrated. 
But, indeed, this Proposition, which makes the loth Defini- 
tion of the nth Book, is not true universally, except in the 
case in which each of the solid angles of the figures is con- 
tained by no more than three plane angles j for in other cases, 
two solid figures may be contained by similar planes of the 
same number and magnitude, and yet be unequal to one ano- 
ther, as shall be made evident in the Notes subjoined to these 
Elements. In like manner, in the Demonstration of the 26th 
Prop, of the i ith Book, it is taken for granted, that those 
solid angles are equal to one another which are contained by 
plain angles of the same number and magnitude, placed ia 
the same order; but neither is this universally true, except 
in the case in which the solid angles are contained by no more 
than three plane angles ; nor of this case is there any Demon- 
stration in the Elements we now have, though it be quite ne- 
cessary there should be one. Now, upon the ioth Definition 
of this Book depend the 25th and 28th Propositions of it ; 
and, upon the 25th and 26th depend other eight, viz* the 27th, 
31st, 32d, 33d, 34th, 36th, 37th, and 40th of the same 
Book; and the 12th of the 1 2th Book depends upon the eighth 
of the same; and this eighth, and the Corollary of Proposition 
17th and Proposition i8th of the I2th Book, depend upon the 
gth D finition of the nth Book, which is not a right defini- 
tion ; because there may be solids contained by the same num- 
ber of similar plane figures, which are not similar to one ano- 
ther, in the true sense of similarity received by geometers ; 
and all these Propositions have, for these reasons, been insuf- 
ficiently demonstrated since Theon's time hitherto. Besides, 
there are several other things, which have nothing of Euclid's 
accuracy, and which plainly shew, that his Elements have 
been much corrupted by unskilful geometers; and though 
these are not so gross as the others now mentioned, they ought 
by no means to remain uncorrected. 

Upon these accounts it appeared necessary, and I hope will 
prove acceptable, to all lovers of accurate reasoning, and of 



PRtFACE. VU 

mathematical learning, to remove such blemishes, and restore 
the principal Books of the Elements to their original accuracy, 
as far as I was able ; especially since these Elements are the 
foundation of a science by which the investigation and disco- 
very of useful truths, at least in mathematical learning, is pro- 
moted as far as the limited powers of the mind allow j and 
which likewise is of the greatest use in the arts both of peace 
and war, to many of which geometry is absolutely necessary. 
This I have endeavoured to do, by taking away the inaccu- 
rate and false reasonings which unskilful editors have put into 
the place of some of the genuine Demonstrations of Euclid, 
who has ever been justly celebrated as the most accurate of 
geometers, and by restoring to him those things which Theon 
or others have suppressed, and which have these many ages 
been buried in oblivion. 

In this edition, Ptolemy's Proposition concerning a pro- 
perty of quadrilateral figures in a circle, is added at the 
end of the sixth Book, Also the Note on the 29th Proposition, 
Book 1st, is altered, and made more explicit, and a more gene- 
ral Demonstration is given, instead of that which was in the 
Note on the lOth Definition of Book nth; besides,, the 
Translation is much amended by the friendly assistance of a 
learned gentleman. 

To which are also added, the Elements of Plane and Sphe- 
fical Trigonometry, which are commonly taught after the 
Elements of Euclid. 



ADVERTISEMENT. 



J- HE favourable reception which former edi- 
tions of Professor Simson^s Elements of Euclid 
have met with from the public^ induced the pro- 
prietors of the work to carry into execution 
every measure most likely to secure and continue 
general approbation. With this view, the pre- 
sent edition has been carefully revised throughout, 
by a very eminent mathematician; for the conve- 
nience of tutors, as well as students, a short 
treatise on the Construction" of the Trigo- 
nometrical Canon has now been inserted, 
from a late celebrated author ; and to this has 
been added, a co?icise Account o/" Logarithms, 
and improved methods of calculating them, by 
the present Savilian Professor of Geometry in 
the Univej'sity of Oxfords 

THE 



THE 



ELEMENTS 



OF 



EUCLID. 



BOOK I. 
DEFINITIONS. 

I. 

A POINT is that which hath no parts, or which hath no book i. 
magnitude. v^^v,^^./ 

J{^ See Notes. 

A line is length without breadth. 

III. 
The extremities of a line are points. 

IV. 
A straight line is that which lies evenly between its extreme 
points. 

V. 
A superficies is that which hath only length and breadth. 

VI. 
The extremities of a superficies are lines. 

VII. 
A plane superficies is that in which any two points being taken, See N. 
the straight line between them lies wholly in chat superficies. 
VIII. 
** A plane angle is the inclination of two lines to one another See N, 
" in a plane, which meet together, but are not in the same 
*' direction." 

IX. 
A plane redlilineal angle is the inclination of two straight 
lines to one another, which meet together, but are not in 
the same straight line. 

B N.B. 



Book I. 



THE ELEMENTS 




N. B. ' When several angles are at one point B, any one 
« of them is expressed by three letters, of which the letter that 

* is at the vertex of the angle, that is, at the point in which 

* the straight lines that contain the angle meet one another, 
' is put between the other two letters, and one of these two is 
' somewhere upon one of those straight lines, and the other 
' upon the other line : Thus the angle which is contained by 

* the straight lines AB,CB, is named the angle ABC, or CB A; 

* that which is contained by AB, BD is named the angle 

* ABD, or DBA ; and that which is contained by BD, CB 
' IS caUed the angle DBC, or CBD ; but, if there be only 
' one angle at a point, it may be expressed by a letter placed 
' at that point ; as the angle at E.' 

When a straight line standing on ano- 
ther straight line makes the adjacent 
angles equal -to one another, each of 
the angles is called a right angle ; 
and the straight line which stands on 

the other is called a perpendicular to 

it. 

XI. 

An obtuse angle is that which is greater than a right angle. 




Xli. 
An acute angle is that which is less than a right angle. 

XIII. 
" A term or boundary is the extremity of any thing." 

A figure is that which is inclosed by one or more boundaries. 



OF EUCLID. 3 

XV. ^^^ 

A circle is a plane figure contained by one line, which is cal- ^^^^"""^ 
led the circumference, and is such that all straight lines 
drawn from a certain point within the figure to the cir- 
cumference, are equal to one another. 




(C 



XVI. 
And this point is called the centre of the circle. 

XVII. 
A diameter of a circle is a straight line drawn through the See N. 
centre, and terminated both ways by the circumference. 

xvm: 

A semicircle is the figure contained by a diameter and the part 
of the circumference cut ofF by the diameter. 

XIX. 
A segment of a circle is the figure contained by a straight 
" line, and the circumference it cuts off." 
XX. 

Reftilineal figures are those which are contained by straight 
lines. 

XXI. 
Trilateral figures, or triangles, by three straight lines. 

XXII. * 

Quadrilateral, by four straight lines. 

XXIII. 
Mutilateral figures, or polygons, by more than four straight 
lines. 

XXIV. 
Of three-sided figures, an equilateral triangle is that which 
has three equal sides. 

XXV. 
An isosceles triangle is that which has only two sides equal. 

B2 



THE ELEMENTS 



Book T. 





XXVI. 
A scalene triangle, is that which, has three unequal sides. 

XXVII. 
A right angled triangle, is that which has a right angle. 

XXVIII. 
An obtuse angled triangle, is that which has an obtuse angle. 




XXIX. 

An acute angled triangle, is that which has threeacu^e angles. 

XXX. 
Of four-fided figures, a square is that which has all its sides 
equal, and all its angles right angles. 



XXXI. 

An oblong, is that which has all its angles right angles, but 

has not all its sides equal. 

XXXII. 
A rhombus, is that which has its sides equal, but its angles 

are not right angles. 




XXXIII. 
iSce N. A rhomboid, is that which has its opposite sides equal to one 
another, but all its sides are not equal, nor its angles right 
angles. 



OF EUCLID. 5 

XXXIV. ^2^ 

All other four-sided figures besides these, are called Trape- ^^^^^^ 

ziums. 

XXXV. 
Parallel straight lines, are such as are in the same plane, and 

which, being produced ever so far both ways, do not meet. 



POSTULATES. 



T ^ 

X-iET it be granted that a straight line may be drawn from 
any one point to any other point. 

II. 
That a terminated straight line may be produced to any length 
in a straight line. 

^ III. 

And that a circle may be described from any centre, at any 
distance from that centre. 

AXIOMS. 

rp I. 

J. HINGS which are equal to the same are equal to one 
another. 

IL ^ - - 

If equals be added to equals, the wholes are equal, '^-i- "- = -^ 

III. 
'If equals be taken from equals, the remainders are equal. 

IV. 
If equals be added to unequals, the wholes are unequal. 

V. 
If equals be taken from unequals, the remainders are unequal. 

VI. 
Things which are double of the same, are equal to one another. 

VII. 
Things which are halves of the same, are equal to one another. 

VIII. 
Magnitudes which coincide with one another, that is, which 
exadly fill the sams space, are equal to one another. 
B3 



) THE ELEMENTS 

Book I. IX. 

*'"*'''^"*''^ The whole is greater than its part. 

X. 
Two straight lines cannot incfose a space. 

XI. 
All right angles are equal to one another. 

XII. 
*' If a straight line meets two straight lines, so as t© make 
the two interior angles on the same side of it taken toge- 
ther less than two right angles, these straight lines being 
continually produced, shall at length meet upon that side 
" on which are the angles which are less than two right 
'' angles. See the notes on Prop. 29. of Book I." 



cc 



OF EUCLID. 




PROPOSITION I. PROBLEM. 

A O describe an equilateral triangle upon a given 
iinite straight line. 

Let AB be the given straight line; it is required to describe 
an equilatecal triangle uf>on it. 

From the centre A, at the dis- 
tance AB, describe* the circle 
BCD, and from the centre B,at 
the distance B/\, describe the' 
circle ACE ; and from the point D 
C, in which the circhs cut one 
another, draw the straight lines'' 
CAjCB to the points A,B; ABC 
shall be an equilateral triangle. 

Because the point A is the centre of the circle BCD, AC is 
equal "^to AB ; and. because the point B is the centre of the 
circle ACt, BC is equal to BA : But it has been proved that 
CA is equal to AB ; therefore CA,CB are each of them equal 
to AB ; but things which are equal to the same are equal to 
one another ^ ; therefore CA is equal to CB ; wherefore CA, 
AB, BC are equal to one another ; and the triangle ABC is 
therefore equilateral, and it is described upon the given straight 
line AB. Which was required to be done. 

PROP. II. PROB. 

£ ROM a given point to draw a straight line.equal 
to a o-iven straight line. 

o o 

Let A be the given point, and BC the given straight line ; it 
is required to draw from the point A a straight line equal toBC 

From the point A to B draw» the 
straight line AB ; and upon it de- 
scribe'' the equilateral triangle 
DAB, and produce"^ the straight 
lines Da, DB, to E and F ; from 
the centre B, at the distance BC, 
describe** the circle CGH, and 
from the centre D, at the distance 
DG, describe the circle GK!L. AL 
shall be equal to BC. 

B 4 Because 



Book I. 



Postu- 



late. 



"IPost 



<= 15 Definj- 

tiOD. 



" lit Axi- 
om. 




M Post. 



1. 



«2Po»t. 



3Fot 



8 THE ELEMENTS 

Bgok r. Because the point B is the centre of the circle CGH, BC is 
•^TdT^ ^^li^l^ ^o BG ; and because Dis the centre of the circle GKL, 
DL is equal to DG, and DA, DB, parts of them, are equal ; 
'3 Ax. therefore the remainder AL is equal to the remainder*^ BG: 
But it has been shown, that BC is equal to BG j wherefore 
AL and BC are each of them equal to BG j and things that 
are equal to the same are equal to one another j therefore the 
straight line AL is equal to BC. Wherefore from the given 
point A a straight line AL has been drawn equal to the given 
straight line BC. Which was to be done. 



PROP. 111. PROB. 



F] 




ROM the greater of two given straight lines to 

cut off a part equal to the less. 
Let AB and C be the two given 

straight lines, whereof AB is the 

greater. It is required to cut off 

from AB, the greater, a part equal 

to C, the less. 
•2.1. From the point A draw* the 

straight line AD equal to C ; and 

from the centre A, and at the dis- 
"»3Post. tance AD, describe'' the circle 

DEF ; and because A is the centre 

of the circle DEFj AE shall be equal to AD ; but the straight 

line C is likewise equal to AD ; whence AE and C are each 

of them equal to AD ; wherefore the straight line AE is equal 
•= 1 Ax. to '^C, and from AB, the greater of two straight lines, a part 

AE has been cut oft equal to C the less. Which was to be 

done. 

PROP. IV. THEOREM. 

J F two triangles have two sides of the one equal to 
two sides of the other, each to each ; and have like- 
wise the angles contained by those sides equal to one 
anotlier; they shall likewise have their bases, ov third 
sides, equal ; and the two triangles shall be equal ;' 
and their other angles shall be e(jual, each to each, 
viz. those to which the equal sides are opposite. 

Let ABC, DEF be two triangles, which have the two sides 

AB, AC equal to the two sides DE, DF, each to each, viz. 

2 AB 



OF EUCLID. 



Book. I. 




ABtoDE,andACtoDF; 
and the angle BAG equal to 
the angle EDP\ the base 
BC shall be equal to the 
base EF ; and the triangle 
ABCto the triangle DEF; 
and the other angles to 
which the equal sides are 
opposite,shall be equal each 
to each, viz. the angle 
ABC to the angle DEP\ 
and the angle ACB to DFE. 

For, if the triangle ABC be applied to DEF, so that the 
point A may be on D, and the straight line A3 upon DE ; 
the point B shall coincide with the point E, because AB is equal 
toDE ; and AB coinciding with DE, AC shall coincide with 
DF, because the angle BAG is equal to the angle EDF ; 
wherefore also the point C shall coincide with the point F, 
because the straight line AC is equal to DF : But the point B 
coincides with the point E ; wherefore the base BC shall coin- 
cide with the base EF, because the point B coinciding with E, 
and C with F, if the base BC does not coincide with the base 
EF, two straight lines would inclose a space, which is impossi- 
blc\ Therefore the base BC shall coincide with the base EF. « 10 Ax. 
and be equal to it. Wherefore the whole triangle ABC shall 
coincide with the whole triangle DEF, and he equal to it ; and 
the other angles of the one shall coincide with the remaining 
angles of the other, and be equal to them, viz. the angle ABC 
to the angle DEF, and the angle ACB to DFE. Theretore, 
if two triangles have two sides of the one equal to two sides 
of the other, each to each, and have likewise the angles con- 
tained by those sides equal to one another, their bases shall 
likewise be equal, and the triangles be equal, and their other 
angles to which the equal sides are opposite shall be equal, 
each to each. Which was to be demonstrated. 

• PROP. V. THEOR. 

1 HE angles at the base of. an Isosceles triangle 
are equal to one another ; and if the equal sides be 
produced, the angles upon the other side of the 
base shall be equal. 

Let ABC be an Isosceles triangle, of which the side AB is 
{ equal 



Book I. 



3.1. 



'>4. 1. 



I Ax. 




THE ELEMENTS 

equal to AC, and let the straight lines AB, AC be produced 
to D and E, the angle ABC shall be equal to the angle ACB, 
and the angle CBD to the angle BCE. 

In BD take any point F, and from AE the greater, cut off 
AG equal* to AF, the less, and join FC, GB. 

Because AF is equal to AG, and AB to AC, the two sides 
FA, AC are equal to the two GA, AB, each to each j and 
they contain the angle FAG common 
to the two triangles AFC, AGB ; 
therefore the base FC is equal '' to 
the base GB, and the triangle AFC^ 
to the triangle AGB j and the remain- 
ing angles of the one are equaP to 
the remaining angles of the other,each 
to each, to which the equal sides are 
opposite ; viz. the angle ACF to the 
angle ABG, and the angle AFC to 
the angle AGB : And because the 
whole AF, is equal to the whole, AG 
of which the parts AB, AC, are-L' 
equal ; the remainder BF shall be 

equal '^ to the remainder CG ; and FC was proved to be equal 
to GB ; therefore the two sides BF, FC are equal to the two 
CG, GB, each to each ; and the angle BFC is equal to the 
angle CGB, and the base BC is common to the two trian- 
gles BFC, CGB J wherefore the triangles are equal'', and 
their remaining angles, each to each, to which the equal sides 
are opposite ; therefore the angle FBC is equal to the angle 
GCB, and the angle BCF to the angle CBG : And, since it 
has been demonstrated, that the whole ^ngle ABG is equal 
to the whole ACF, the parts of which, the angles CBG, BCF 
are also equal 5 the remaining angle ABC is therefore equal 
to the remaining angle ACB, which are the angles at the 
base of the triangle ABC: And it has also been proved that 
the angle FBC is equal to the angle ^CB, which are the 
angles upon the other side of the base. Therefore the "angles 
at the base. Sec. Q^E.D. 

Corollary. Hence every equilateral triangle is also 
equiangular. 

PROP. VI. THEOR. , 

X F two angles of a tjj^nglc be eijual to one ano- 
ther, the sides also which subtend, or arc opposite 
tOy the equal angles, shall be equal to one another. 

Let 




b-i. 1. 



OF EUCLID. II 

Let ABC be a triangle having the angle ABC equal to the ^°°^ ^• 
angle ACB ; the side AB is also equal to the side AC. ^-^/^«' 

For, if AB be not equal to AC, one of them is greater than 
the other : Let AB be the greater ; and from it cut* ofF DB » 3- 1. 
equal to AC, the less, and join DC ; there- 
fore, because in the triangles DBC, ACB, 
DB is equal to AC, and BC common to 
both the two sides, DB, BC are equal to 
the two AC, CB each to each ; and the 
angle DBC is equal to the angle ACB ; 
therefore the base DC is equal to the base 
AB, and the triangle DBC is equal to the 
triangle'' ACB, the less to the greater ; 
which is absurd. Therefore AB is not L" 
unequal to AC, that is, it is equal to it. 
Wherefore, if two angles, &c. Q. E. D. 

Cor. Hence every equiangular triangle is also equilateral. 

PROP. Vn. THEOR. 

L PON the same base, and on the same side of it, s** n, 
there cannot be two triangles that have their sides 
which are terminated in one extremity of the base 
equal to one another, and likewise those which are 
terminated in .tlie other extremity. 

If it be possible, let there be two triangles ACB, ADB, up- 
on the same base AB, and upon the same side of it, which have 
their sides CA, DA, terminated in the extremity A of the 
base equal to one another, and liice- 
wise their sides CB, DB,that are ter- 
minated in B. 

Join CD ; then, in the case in 
which the vertex o^ each of the tri- 
angles is without the other triangle, 

because AC is equal to AD, the , ^ -. , ,^ ^ 

angle ACD is equal* to the angle 
ADC : But the an^le ACD is greater 
than the angle BCD; therefore the 
angle ADC is greater also than BCD; 

much more then is the angle BDC greater than the angle 
BCD. Again, because CB is equal to DB, the angle BDC 
is equal* to the angle BCD ; but it has been demonstrated to 
be greater than it j which is impossible. 

But 




12 



THE ELEMENTS 



3. J. 




Book 1. gut if one of the vertices, as D, be within the other triangle 

^'^^^'"^ ACB ; produce AC, AD to E, F •, 
therefore, because AC is equal to AD 
in the triangle ACD, the angles ECD, 
FDC upon the other side of the bas'i 
CD are equal^ to one another, but 
the angle ECD is greater than the 
angle BCD ; wherefore the angle 
FDC is likewise greater than BCD ; 
much more then is the angle BDC 
greater than the angle BCD. Again, 
because CB is equal to DB, the angle 

BDC is equal "^ to the angle BCD j but BDC has been proved 
to be greater than the same BCD ; which is unpossible. The 
case in which the vertex of one triangle is upon a side of the 
other, needs no demonstration. 

Therefore, upon the same base, and on the same side of it, 
there cannot be two triangles that have their sides which are 
terminated in one extremity of the base equal to one another, 
and likewise those which are terminated in the other extre- 
mity. Q.E.D. 

PROP. VIII. THEOR. 

If two triangles have two sides of the one equal to 
two sides of the other, each to each, and have like- 
wise their bases equal ; the angle which is contain- 
ed by the two sides of the one shall be equal to the 
angle contained by the two sides equal to them, of 
the other. 

Let ABC, DEF be two triangles, having the two sides AB, 
AC, equal to the two sides DE, DF, each to each, viz. AB 
to DE, and AC a ^^ P 

to DF ; and also 
the baseBC equal 
to the base EF. 
The angle BAG 
is equal to the 
angle EDF. 

For, if the tri- 
angle ABC be 
applied to DEF, 

so that the point B be on E, and the straight line BC upon 
EF ; the point C shall also coincide with the point F. Because 

BC 




OF EUCLID. 



^3 



BC is equal to EF ; therefore BC coinciding with EF ; BA BookL 
and AC shall coincide with ED and DF ; for, it" the base BC ^""^'^■^ 
coincides with the base EF, but the sides BA, CA do not 
coincide with the sides ED, FD, but have a different situation 
as EG, FG, then, upon the same base EF, and upon the same 
side of" it, there can be two triangles that have their sides which 
are terminated in one extremity ot the base equal to one ano- 
ther, and likewise their sides terminated in the other extremity : 
But this is impossible* ; therefore, if the base BC coincides* '• '• 
with the base EF, the sides BA, AC cannot but coincide with 
the sides ED, DF ; wherefore likewise the angle BAC coin- 
cides with the angle EDF, and is equal *• to it. Therefore if ^ ^^ 
two triangles, &c. Q^E, D. 

PROP. IX. PROB. 

A O bisect a given rectilineal angle, that is, to di- 
vide it into two equal angles. 

Let BAC be the given rectilineal angle, it is required to 
bisect it. 

Take any point D in AB, and from AC cut ' off AE equal • 3- i- 
to AD; join DE, and upon it describe** •> 1. 1. 

an equilateral tiiangle DEF; then join 
AF ; the straight line AF blsedls the angle 
BAC. 

Because AD is equal to AE, and AF is 
common to the two triangles DAF, EAFj 
the two sides DA, AF, are equal to the 
tWo sides EA, AF, each to each ; and the < 
base DF is equal to the base EF ; there--! 
fore the angle DAF is equal<^ to the angle 
EAF ; wherefore the given rectilineal 
angle BAC is bisefted by the straight line AF. Which was 
to be done. 

PROP. X. PROB. 




•=8. 1. 



1 O bisect a given finite straight Hne, that is, to 
divide it into two equal parts. 

Let AB be the given straight line ; it is required to divide 
4t into two equal parts. 

Describe* upon it an equilateral triangle ABC, and bise<Sl * ^ 
''the angle ACB by the straight line CD. AB is cut into two 
equal parts in the point D. Because 



1. 

9. 1. 



Book I. 



■4,1. 



THE ELEMENTS 

Because AC is equal to CB, and CD 
common to the two triangles ACD, 
BCD ; the two sides AC, CD are equal 
to BCj CD, each to each ; and the 
angle ACD is equal to the angle BCD ; 
therefore the base AD is equal to the 
base'^ DB, and the straight line AB is 
divided into two equal parts in the point V 
D. Which was to be done. 




SeeN. 



» .3. 1 . 
»1. 1. 



«8. 1. 



<»10Def. 



PROP. XI. PROB. 

X O draw a straight line at right angles to a given 
straight line, from a given point in the same. 

• Let AB be a given straight line, and C a point given in 
it ; it is required to draw a straight line from the point C at 
right angles to AB. 

Take any point D in AC, and ^ make CE equal to CD, and 
upon DE describe ^ the equi- 
lateral triangle DFE, and join 
FC, the straight line FC drawn 
from the given point C is at 
right angles to the given 
straight line AB.. 

Because DC is equal to CE, 
and FC common to the two ^ 
triangles DCF,ECFj the two 
sides DC, CF, are equal to the two EC,CF, each to each ; and~ 
the base DF is equal to the base EF j therefore the angle DCF 
is equal "^ to the angle ECF ; and they are adjacent angles. 
But, when the adjacent angles which one straight line makes 
with another straight line are equal to one anothcr,each of them 
is called a right'' angle j therefore each of the angles DCF, 
ECF, is a right angle. Wherefore, from the given point C, 
in the given straight line AB, FC has been drawn at right 
angles to AB. Which was to be done. 

Cor. By help of this problem, it may be demonstrated, 
that two straight lines cannot have a common segment. 

If it be possible, let the two straight lines ABC, ABD have 
the segment AB common to both of them. Fiom the point B 
draw BE at right angles to AB j and because ABC is a straight 

line, 




OF EUCLID. 



line, the angle CBE is equal * to 
the angle EBA ; in the same 
manner, because ABD is a 
straight line, the angle DBE is 
equal to the angle EBA ; where- 
fore the angle DBE is equal to 
the angle CBE, the less to the 
greater ; which is impossible ; 
therefore two straight lines can- 
not have a common segment. 



E 



A 




D 



PROP. XII. PROB. 




X O draw a straight line perpendicular to a given 
straight line of an unlimited length, from a given 
point witiiout it. 

Let AB be the given straight line, which may be produced 
to any length both ways, and let C be a point without it. It is 
required to draw a straight line 
perpendicular to AB from the 
point C, 

Take any point D upon the 
other side of AB, and from the 
centre C, at the distance CD, 
describe'' the circle EGF meet- 
ing AB in FG ; and bisect "^FG 
in H, and join CF, CH, CG ; 

the straight line CH, drav/n from the given point C, is per- 
pendicular to the given straight line AB. 

Because FH is equal to HG, and HC common to the two 
triangles FHC, GHC, the two sides FH, HC are equal to the 
two GH, HC, each to each ; and the baseCF is equaH to the « 15 Dcf. 
base CG ; therefore the angle CKF is equal • to the angle j 
CHG ; and they are adjacent angles j but when a straight line 
standing on a straight line makes the adjacent angles equal to 
one another, each of them is a right angle ; and the straight line 
which stands upon the other is calbd a perpendicular to it ; 
therefore from the given point C a perpendicular CH has been 
drawn to the giYen straight line AB. Which was. to be done. 

PROP. XIII. THEOR. 

X HE angles which one straight line makes with 
another upon the one side of it, are either two right 
angles, or are together equal to two right angles. 

Let 



s, 1. 




• Def. 10. 
«>11. 1. 



«2Ax. 



••l Ax. 



THE ELExMENTS 

Let the straight line AB make with CD, upon one side of 
it, the angles CBA, ABD ; these are either two right angles, 
or are together equal to two right angles. 

For if the angle CBA be equal to ABD, each of them is a 

E 

A 



D 



A 




B 



right ^ angle ; but, if not, from the point B draw BE at right 
angles ^ to CD ; therefore the angles CBE, EBD are two right 
angles* ; and because CBE is equal to the two angles CBA; 
ABE together, add the angle EBD to each of these equals j 
therefore the angles CBE, EBD are equal "^ to the three an- 
gles CBA, ABE, EBD. Again, because the angle DBA is 
equal to the two angles DBE, EBA, add to these equals the 
angle ABC, therefore the angles DBA, ABC are equal to the 
three angles DBE, EBA, ABC ; but the angles CBE, EBD 
have been demonstrated to be equal to the same three angles; 
and things that are equal to the same are equal** to one another ; 
therefore the angle CBE, EBD are equal to the angles DBA, 
ABC ; but CBE, EBD are two right angles ; therefore DBA, 
ABC are together equal to two right angles. Wherefore 
when a straight line, &c. Q. E. D. 

PROP. XIV. THEOR. 

XF, at a point in a straight line, two other straight 
lines, upon the opposite sides of it, make the adja- 
cent angles, together equal to two right angles^ 
these two straight lines shall be in one and the 
same straight line. ' 

At the point B in th:; straight .\ 

line AB,let the two straight lines 
BC, BD upon the opposite sides 
of AB, make the adjacent an- 
gles ABC, ABD equal together 
to two right angles, BD is in the 
same straight line with CB. 

For, ifBDbenot in the sameC- 
straight line with CB, let BE be 




B 



D 



in 



OF EUCLID. 17 

in the same straight line with it ; therefore, because the straight BookL 
line AB makes angles with the straight line CBE, upon one 
side of it, the angles ABC, ABE are together equal* to two »13. 1. 
right angles; but the angles ABC, ABD are li ice wise toge- 
ther equal to two right angles ; therefore the angles CBA, ABE 
are equal to the angles CBA, ABD : Take away the common 
angle AEC, the remaining angle ABE is equaP to the re- "3. As. 
maining angle ABD, the less to the greater, which is impos- 
sible ; therefore BE is not in the same straight line with BC. 
And, in like manner, it may be demonstrated, that no other 
can be in the same straigh: line with it but BD, which there- 
fore is in the same straight line with CB. Wherefore, if at 
a point, &c. Q^ E. D. 

PROP. XV. THEOR. 



1 F two straight lines cut one another, the vertical, 
or opposite, angles shall be equal. 

Let the two straight lines AB, CD, cut one another in the 
point E ; the angle AEC shall be equal to the angle DEB, 
and CEB to AED. 

Because the straight line AE 
makes with CD the angles CE A, ^ 
AED, these angles are together 
equal * to two right angles. — 
Again, because the straight line 
DE makes with AB the angles 
AED, DEB, these also are to- 
gether equal* to two right an- 
gles; and CEA, AED, have been 
demonstrated to be equal to two right angles ; wherefore the 
angles CEA, AED, are equal to the angles AED, DEB. Take 
away the common angle AED, and the remaining angle CEA 
is equal'' to the remaining angle DEB. In the same manner "3- -Ax, 
it can be demonstrated, that the angles CEB, AED are equal. 
Therefore, if two straight lines, &c. Q. E. D. 

Cor. I. From this it is manifest, that, if two straiglit lines 
cut one another, the angles they make at the point where they 
cut, are together equal to four right angles. 

Cor. 2. And consequently that all the angles made by any 
number of lines meeting in one point, are together equal to 
four right angles. 

C 




M3. I. 



iS 



THE ELEMENTS 



Book I. 



»10. 1. 



*lb. 1. 



*. 1. 



'IS. 1. 



PROP. XVI. THEOR. 

If one side of a triangle be produced, the exterior 
angle is greater than either of the interior opposite 
angles. 

Let ABC be a triangle, and let its side BC be produced to 
D, the exterior angle ACD is greater than either of the inte- 
rior opposite angles CBA, BAC. 

Bisect* AC in E, join BE A 

and produce it to F, and 
make EF equal to BE j join 
also FC, and produce AC to 
G. 

Because AE is equal to 
EC, and BE to EF ; AE, 
EB are equal to CE, EF, 
each to each j and the angle H 
AEB, is equal** to the angle 
CEF, because they are op- 
posite vertical angles ; there- . 
fore the base AB is equa^ to 
the base CF, and the triangle 

AEB to the triangle CEF, and the remaining angles to the 
remaining angles, each to each, to which the equal sides are 
opposite i wherefore the angle BAE is equal to the angle 
ECF i but the aiigle ECD is greater than the angle ECF ; 
therefore the angle ACD is greater than BAE : In the same 
manner, if the side BC be bisected, it may be demonstrated 
that the angle BCG, that is"*, the angle ACD, is greater than 
the angle ABC. Therefore, if one side, &c. Q; E. D. 




'\6. 1. 



A 



PROP. XVIL THEOR. 



NY two angles of a triangle are together less 
than two rioht an":les. 

Let ABC be any triangle ; any 
two of its angles together are 
less than two right angles. 

Produce BC to Dj and be- 
cause ACD is the exterior ansle 
of the triangle ABC, ACDTis 
greater' than the interior and 
oppo^te angle ABC ; to each of 




OF EUCLID. 



'9 




•3. 1. 



" 1&. 1. 



these add the angle ACB ; therefore the angles ACD, ACB Boor I. 
are greather than the angles ABC, ACB ; but ACD, ACB '"^^''^^ 
are together equal** to two right angles; therefore the angles » 13. 1. 

ABC, BCA are less than two right angles. In like manner, 
it may be demonstrated, that BAC, ACB, as also CAB, ABC, 
are less than two right angles. Therefore any two angles, 
&c. Q, E. D. 

PROP. XVIII. THEOR. 

JL HE greater side of every triangle is opposite to 
the greater angle. 

Let ABC be a triangle, of 
which the side AC is greater 
than the side AB ; the angle 
ABC is also greater than the 
angle BCA. 

Because AC is greater than 
AB, make » AD equal to AB, 
and join BD j and because ADB 
is the exterior angle of the tri- 
angle BDC, it is greater ^ than 
the interior and opposite angle DCB j but ADB is equal^ to «5. 1. 

ABD, because the side AB is equal to the side AD ; therefore 
the angle ABD is likewise greater than the angle ACB. 
Wherefore much more is the angle ABC greater than ACB. 
Therefore the greater side, &c. Q^ E. D. 

PROP. XIX. THEOR. 

X HE greater angle of every triangle is subtended 
by the greater side, or has the greater side opposite 
to it. 

Let ABC be a triangle, of which the angle ABC is greater 
than the angle BCA ; the side AC is likewise greater than the 
side AB. 

For, if it be not greater, AC, 
must either be equal to AB, or 
less than it j it is not equal, be- 
cause then the angle ABC would 
be equal * to the angle ACB ; 
but It is not J therefore AC is 
not equal to AB j neither is it 

less ; became ikca the. angle _ 

Cl ABC 




^. I. 



20 THE ELEMENTS 

Book i,^ ABC would be less '' than the angle ACB i but it is not i 
" ~ ' therefore the side AC is not less than AB ; and it has been 
shewn that it is not equal to AB ; therefore AC is greater than 
AB. Wherefore the greater angle, &c. (^ E. D. 



" 18. 1. 




PROP. XX. THEOR. 

SeeN. Any two sidcs of a triangle are together greater 
than the third side. 

Let ABC be a triangle j any two sides of it together are 
greater than the third side, viz. the sides BA, AC greater than 
the side BC ; and AB, BC greater than AC ; and BC, CA 
greater than AB. 

Produce BA to the point D, 

•3. J- and make^ AD equal to AC j 
and join DC. 

Because DA is equal to AC, 
the angle ADC is likewise equal 

»- 5. 1. b to ACD ; but the angle BCD 
is greater than the angle ACD ; 
therefore theangle BCD is great- 
er than the angle ADC ; and because the angle BCD of the 
triangle DCB is greater than its angle BDC, and that the 

« 19 1. greater*^ side is opposite to the greater angle; therefore the 
side DB is greater than the side BC ; but DB is equal to BA 
and AC ; therefore the sides BA, AC are greater than BC. 
In the same manner it may be demonstrated, that the sides 
AB, BC are greater than CA, and BC, CA greater than AB. 
Therefore any two sides, &c. Q^ E. D. 

PROP. XXL THEOR. 

5eeN. If, from, the ends of the side of a triangle, there be 
drawn two straight lines to a point within the trian- 
gle, these shall be less than the other two sides of 
the triangle, but shall contain a greater angle. 

Let the two straight lines BD, CD be drawn from B, C, 
the ends of the side BC of the triangle ABC, to the point D 
within it? BD and DC are less than the other two sides BA, 
AC of the triangle, but contain an angle BDC greater thao 
the angle BAC. 

Produce BD to E j and because two sides of a triangle are 
greater than tlje third side, the two sides BA, AE of the tri- 

2 suigle 



OF EUCLID. 



21 




angle ABE are greater than BE. To each of these add EC j BookL 

therefore the sides BA, AC ^•^n'*^ 

are greater than BE, EC : A- 

gain, because the two sides CE, 

ED of the triangle CED are 

greater than CD, add DB to 

each of these ; therefore the 

sides CE, EB are greater than 

CD, DB ; but it has been 

shewn that B A, AC are greater 

than BE, EC, much more then 

are BA, AC greater than BD, DC. 

Again, because the exterior angle of a triangle is greater 
than the interior and opposite angle, the exterior angle BDC 
of the triangle CDE is greater than CED ; for the same rea- 
son, the exterior angle CEB of the triangle ABE is greater 
than BAC ; and it has been demonstrated that the angle BDC 
is greater than the angle CEB ; much more then is the angle 
BDC greater than the angle BAC. Therefore, if from the 
ends of, &c. Q E. D. 

PROP. XXII. PROB. 

X O make a triangle of which the sides shall besceN. 

equal to three given straight lines, but any two 
whatever of these must be greater than the third.* »20. i 

Let A, B, C be the three given straight lines, of which any 
two whatever are greater than the third, viz. A and B greater 
than C ; A and C greater than B j and B and C than A. It 
is required to make a triangle of which the side shall be equal 
to A, B, C, each to each. 

Take a straight line DE terminated at the point D, but un- 
limited towards E, and 
make * DF equal to A, 
FG to B, and GH equal 
to C; and from the cen- 
tre F, at the distance ^ 
FD, describe ''the circle'^ 
DKL ; and from the cen- 
tre G, at the distance 
GH, describe ^ another 
circle HLK j and join 
KF, KG; the triangle 
KFG has its sides equal to the three straight lines A, B, C. 

3ecaus« the point F is the centre of the circle DKL, FD is 
C 3 «^ 




«> 3. Past. 



21 THE ELEMENTS. 

Book t. equal ' to FK i but FD is equal to the straight line A ; there- 
^"^S^viM fo''c FK *5 equal to A : Again, because G is the centre of the 
' circle LKH, GH is equal " to GK -, but GPl is equal to C j 
therefore also GK is equal to C ; and FG is equal to B ; there- 
fore the three straight lines KF, FG, GK, are equal to the 
three A, B, C : And therefore the triangle KFG has its 
three sides KF, FG, GK. equal to the three given straight 
lines, A, B, C. Which was to be done. 

PROP. XXIII. PROB. 

jtVT a given point in a given straight line, to make 
a rectilineal angle equal to a given rectilineal angle. 

Let AB be the given straight line, and A the given point 
in it, and DCE the given rectilineal angle ; it is required to 
make an angle at the 

given point A in the q a 

given straight line AB, 
that shall be equal to the 
given rectilineal angle 
DCE. 

Take in CD, CE 
any points D, E, and 
"22. 1. join DE ; and make* 
the triangle AFG, the 
sides of which shall be 
eqiial to the three 

straight lines CD, DE, EC, so that CD be equal to AF, 
CE to AG, and DE to FG ; and because DC, CE are equal 
to F A, AG, each to each, and the base DE to the base FG j 
^^- ■ • the angle DCE is equal ^ to the angle FAG. Therefore, at 
the given point A in the given straight line AB, the angle 
FAG is made equal to the given rectilineal angle DCE. 
Which was to be done. 

PROP. XXIV. THEOR. 

SeeN. 1 F two triangles have two sides of the one equal to 
two sides of the other, each to each, but the angle 
contained by the two sides of one of them greater 
than the angle contained b}^ the two sides equal to 
them, of the other; the base of that which has the 
greater angle shall be greater than the base of the 
. other. 

Let 






OF EUCLID. 2j 

Let ABC, DEF be two triangles which have the two sides ^0°"^ I- 
AB, AC equal to the two DE, DF, each to each, viz. AB ^^^^'^ 
equal to DE, and AC to DF ; but the angle BAC greater than 
the angle EDF ; the base BC is also greater than the base EF. 

Of the two sides DE, DF, let DE be the side which is not 
greater than the other, and at the point D, in the straight line 
DE, make * the angle EDG equal to the angle BAC j and * 23. l. 
make, DG equal ^ to AC or DF, and join EG, GF. ^ 

Because AB is equal to DE, and AC to DG, the two sides 
BA, AC are equal to the two ED, DG, each to each, and the 
angle BAC is equal 

to the angle EDG ; ^ j) 

therefore the baseN 
BC is "^ equal to thel 

base EG; and be-1 \ 1 W * ^' 

cause DG is equal 
to DF, the angle 
DFG is equal'' to the 
angle DGF ; but 
the angle DGF is 
greater than the an- 
gle EGF ; therefore 

the angle DFG is greater than EGF ; and much more is the an- 
gle EFG greater than the angle EGF j and because the angle 
EFG of the triangle EFG is greater than its angle EGF, and 
that the greater = side is opposite to the greater angle ; the side ' l?' 
EG is therefore greater than the side EF ; but EG is equal to 
BC ; and therefore also BC is greater than EF. Therefore, 
if two triangles, &c. Q^ E. D. 

PROP. XXV. THEOR. 

If two triangles have two sides of the one equal to 
two side^ of the other, each to each, but the base 
of the one greater than the base of the other ; the 
angle also contained by the sides of that which baa 
the greater base, shall be greater than the angle 
contained by the sides equal to them of the other. 

Let ABC, DEF be two triangles which have the two sides 
AB, AC equal to the two sides DE, DF, each to each, viz. 
AB equal to DE, and AC to DF ; but the base CB greater 
than the base EF j the angle BAC is likewise greater than the - 
angle EDF. 

C 4 For, 



24 

Book I, 



=■4. 1 



'24. 1 




THE ELEMENTS 

For, if it be not greater, it must either be equal to it,or less j 
but the angle BAC is not equal to the angle EDF, because 
then the base BC 
would beequal^toEF: jV 
but it is not ; there- 
fore the angle BAC is 
not equal to the angle 
EDF ; neither is it 
less ; because then the 
base BC would be less 
• ^ than the base EF ; 
but it Is not ; there- 
fore the angle BAC 

is not less than the angle EDF ; and it was shewn that it is 
not equal to it ; therefore the angle BAC is greater than the 
angle EDF. Wherefore, if two triangles, &c. Q^ E. D. 

PROP. XXVI. THEOR. 

J. F two triangles have two angles of one equal to 
two angles of the other, each to each ; and one side 
equal to one side, viz. either the sides adjacent to 
the equal angles, or the sides opposite to equal an- 
gles in each ; then shall the other sides he equal, 
each to each ; . and also the third angle of the one 
to the third angle of the other. 

Let ABC, DEF be two triangles which have the angles 
ABC, BCA equal to the angles DEF, EFD, viz. ABC to 
DEF, and BCA to EFD ; also one side equal to one side ; 
and first let those sides be equal which are adjacent to the an- 
gles that are equal in the two triangles ; viz. BC to EF ; 
the other sides . 1^ 

shall be equal, each ■ ■ 

to each, viz. AB 
to DE, and AC 
to DFj and the 
third angle BAC 
to lie third angle 

EDF. 

For, if A B be not B ^ C li V 

equal toDE, one of them must be the greater. Let AB be 
the greaterof thetwo, and make BG equal to DE, and joinGCj 
.therefore, because BG is equal to DE, and BC to EF, the two 

sides 





OF EUCLID. 



as 



sides GB, BC are equal to the two DE, EF, each to each ; and ^°°^ ^• 
the angle GBC is equal to the angle DEF ; therefore the base ^""'^'''^^ 
GC is equal* to the base DF, and the triangle GBC to the tri- a 4, 1. 
aqgle DEF, and the other angles to the other angles, each to 
each, to which the equal sides are opposite ; therefore the angle 
GCH is equal to the angle DFE; but DFE is, by the hypo- 
thesis, equal to the an;;le BCA j wherefore also the angle BCG 
is equal to the angle BCA, the less to the greater, which is im- 
possible: therefore AB is not unequal to DE,that is, it is equal 
to it ; and BC is equal to EF ; therefore the two AB, BC are 
equal to the two DE, EF, each to each j and the angle ABC is 
equal to the angle DEF ; the base therefore AC is equal* to 
the base DF, and the third angle B AC to the third angle EDF. 

Next, let the sides 
which are opposite to 

equal angles in each^ *^ 

triangle be equal to 
one another, viz. AB 
to DE ; likewise in 
this case, the other 
sides shall be equal, 
AC to DF, and BC 
to EF j and also the 
third angle BAC to 
the third EDF. 

For, if BC be not equal to EF, let BC be the greater of 
them, and make BH equal to EF, and join AH ; and because 
BH is equal to EF, and AB to DE j the two AB, BH are equal 
to the two DE, EF, each to each ; and they contain equal an- 
gles ; therefore the base AH is equal to the baseDF, and the 
triangle ABH to the triangle DEF, and the other angles shall 
be equal, each to each, to which the equal sides are opposite ; 
therefore the angle BHA is equal to the angle EFD ; but EFD 
is equal to the angle BCA ; therefore also the angle BHA is 
equal to the angle BCA, that is, the exterior angle BHA of 
the triangle AHC is equal to its interior and opposite angle 
BCA ; which is impossible'' ; wherefore BC is not unequal to" 16. 1. 
EF, that is, it is equal to it ; and AB is equal to DE ; therefore 
the two, AB, BC are equal to the two DE, EF, each to each ; 
and they contain equal angles j wherefore the base AC is equal 
to the base DF, and the third angle Bx^C to the third angle 
J^DF. Therefore, if two triangles, Sec. Q.E.D, 





•6 THE ELEMENTS 

Book t. PROP. XXVII. THEOR. 

If a straight line falling upon two other straight 
lines makes the alternate angles equal to one ano- 
ther, these two straight lines shall be parallel. 

Let the straight line E F, which falls upon the two straight 
lines AB, CD make the alternate angles AEF, EFD equal to 
one another; AB is parallel to CD. 

For, if it be not parallel, AB and CD being produced shall 
meet either towards B, D, or towards A, C : let them be pro- 
duced and meet towards B, D in the point G ; therefore GEF 

» 16. 1. is a triangle, and its exterior angle AEF is greater* than the 
interior and opposite angle 
EFG ; but it is also equal to 
it, which is impossible; there- 
fore AB and CD being pro- 
duced do not meet towards 
B, D. In like manner it 
may be demonstrated, that 
they do not meet towards A, 
C ; but those straight lines 
which meet neither way, though produced ever so far, are 

*55Pef. parajlel'' to one another. AB therefore is parallel to CD. 
Wherefore, if a straight line, &c. Q. E. D. 

PROP. XXVm. THEOR. 

jLF a straight line falling upon two other straight 
lines makes the exterior angle equ^l to the interior 
and opposite upon the same side of the line ; or 
makes the interior angles upon the same sidfc toge- 
ther equal to two Tight angles ; the two straight 
lines shall be parallel to one another. 

LetthestraightlineEFjWhich r, 
falls upon the two straight lines \^ 
AB,CD,maketheexterior angle \ 
EGB equal to the interior and « \(t 
opposite angle GHD upon the ' 
same side ; or make the interior 
angles on the same side BGH,^._ 
GHDtogether equal to two right 
angles ; AB is parallel to CD. 

Because the angle EGB is e- 
qual to the angle GHD, and lilie 

angle 




OF EUCLID. 



27 



angle EGB equal* to the angle AGH, the angle AGH is Boos i. 
equal to the angle GHD ; and they are the alternate angles ; i^^f"^ 
therefore AB is parallel^ to CD. Again, becaufe the angles "27. ]. 
BGH, GHD are equal<^ to right angles ; and that AGH, ' By Hip. 
BGH are alfo equal^ to two right angles ; the angles AGH," '^' ^• 
BGH are equal to the two angles BGH, GHD : Take away 
the common angle BGH ; therefore the remaining angle AGH 
is equal to the remaining angle GHD ; and they are alternate 
angles; therefore AB is parallel to CD. Wherefore if a 
ftraight line, &c. Q. E. D. 

PROP. XXIX. THEOR. 

If a straight line fall upon two parallel straight n^^^;^„ 
lines, it makes the alternate angles equal to one '!»'* p™p^ 
another; and the exterior angle equal to the inte- 
rior and opposite upon the same side ; and likewise 
the two interior angles upon the same side together 
equal to two right angles. 

Let the ftraight line EF fall upon the parallel ftraight lines 
AB, CD -J the alternate angles AGH, GHD are equal to one 
another ; and the exterior angle EGB is equal to the interior 
and oppofite, upon the fame fide, 
GHD J and the twointerior angles 
BGH, GHD upon the fame fide, 
are together equal to two right 
angles. 

For, if AGH be not equal to 
GHD, one of them muft be greater 
than the other ; let AGH be the 
greater; and becaufe the angle AGH 
is greater than the angle GHD, add 
to each of them the angle BGH ; therefore the angles AGH, 
BGH are greater than the angles BGH, GHD ; but the angles 
AGH, BGH are equal-* to two right angles ; therefore the » 13. 1. 
angles BGH, GHD are lefs than two right angles ; but those 
ftraight lines which, with another ftraight line falling upon them, 
make the interior angles on the fame fide lefe than two right 
angles, do meet* together if continually produced ; therefore * 12. Ax. 
the ftraight lines AB, CD, if produced far enough, ftiall meet ; ^^ ''"^ 
but they never meet, fmce they are parallel by the hvpothefis; thi) propo- 
thereforethe angle AGH is not unequal to the angle GHD, that '''''°''- 
is, it is equal to it ; but the angle AGH is equal"* to the angle b 15 j, 
EGB j therefore likewife EGB is equal to GHD ; add to each 

of 





THE ELEMENTS 

of thefe the angle BGH ; therefore the angles EGB, BGH are 
('^^i]*^ equal to the angles BGH^ GHD ; but EGB, BGH are equal'^ 
to two right angles; therefore alfo BGH, GHD are equal to 
two right angles. Wherefore, if a ftraight, Sec. Q^ E. D. 

PROP. XXX. THEOR. 

oTRAIGHT lines which are parallel to the same 

straight line are parallel to one another. 

Let AB, CD be each of them parallel to EFj AB is alfo 

parallel to CD. 

Let the ftraight line GHK cut AB, EF, CD i and becaufe 

GHK cuts the parallel ftraight 

lines AB, EF, the angle AGH 
•J9. X. is equal* to the angle GHF. 

Again, becaufe the ftraight line 

GK. cuts the parallel ftraight lines 

EF, CD, the angle GHF is 

equal* jto the angle GKD ; and 

it was fliewn that the angle 

AGK is equal to the angle 

GHF ; therefore alfo AGK is 

equal to GKD j and they are 
*2?.l. alternate angles j therefore'AB is parallel'' to CD. . Where- 
fore ftraight lines, kc. Q^ E. D. 




PROP. XXXI. PROB. 



T. 



O draw a straight line through a given point pa- 
rallel to a given straight line. 

■ Let A be the given point, and BC the given ftraight line •, 
it is required to draw a ftraight line 

through the point A, parallel to the {^^ ^ K 

ftraight line BC. '^ 

In BC take any point D, and join 
AD} and at the point A, in the 

f 23. 1. ftraight line AD, make* the angle tj ry 

DAE equal to the angle ADCj and 
produce the ftraight line EA to F. 

Becaufe the ftraight line AD, which meets the two ftraight 

lines BC, EF, makes the alternate angles EAD, ADC equal 

>^, 1. to one another, EF is parallel'^ to BC. Therefore the ftraight 

line 



OFEUCLID. 29 

line E AF is 3rawn through the given point A parallel to the ^°°^ '• 
given ftraighc line BC. XVhich was to be done. ^ 

PROP. XXXII. THEOR. 

If a side of any triangle be produced, the exterior 
angle is equal to tlie two interior and opposite an- 
gles; and the tliree interior angles of every triangle 
are equal to two right angles. 

Let ABC be a triangle, and let one of its fides BC be 
produced co D ; the exterior angle ACD is equal to the two 
interior and oppofite angles CAB, ABC, and the three inte- 
rior angles of the triangle, viz. ABC, BCA, CAB, are toge- 
ther equal to two right angles, 

ThroughthepointCdraw 
CE parallel* to the ftrai^ht \ »3U 1. 

line AB ; and becaufe AB is 
parallel toCE,and AC meets 
them, the alternate angles 

B AC, ACE are equals A- y^ \ y^ bjg. 1. 

gain, becaufe AB is parallel ^ 
to CE, and BD falls upon '^ 
them,theexteriorangleECD 
is equal to the interior and 

oppofite angle ABC ; but the angle ACE was fhewn to be 
equal to the angle BAC ; therefore the whole exterior angl^ 
ACD is equal to the two interior and oppofite angles CAB, 
ABC ; to thefe equals add the angle ACB, and the angles ACD, 
ACB are equal to the three angles CB A, BAC, ACB ; but the 
angles ACD, ACB, are equal' to two right angles ; therefore c 13. i 
alfo the angles CBA, BAC, ACB are equal to two right an- 
gles. Wherefore if a fide of a triangle, &c. Q^ E. D. 

Cor. I. All the interior angles 
of any reflilineal figure, together 
with four right angles, are equal to ^ 

twice as many right angles as the ^ 

figure has fides. 

For any reftilineal figure ABCDE 

can be divided into as many triangles 

as the figure has fides, by drawing 

ftraight lines from a point F within 

the figure to each of its angles. 

And, by the preceding propofition, 

all 





3» 



THE ELEMENTS 




Book J. a) J t^g angles of these triangles are equal to twice as many 
'"^'^'^'^''^ right apgles as there are triangles, that is, as there are sides 
of the figure ; and the same angles are equal to the angles of 
the figure, together with the angles at the point F, which is 
» 2 Cor. the common vertex of the triangles : that is", together with 
^•'- ^- four right angles. Therefore all.the angles of the figure, to- 
gether with four right angles, are equal to twice as many right 
angles as the figure has sides. 

Cor. 2. All the exterior angles of any rectilineal figure are 
together equal to four rightangles. 
Because every interior angle 
ABC, with its adjacent exterior 
t-is. 1. ABD, is equal ^ to two right 
angles; therefore all the interior 
together with all the exterior 
angles of the figure, are equal to 
twice as inany tvj:\\t angles as 
there are sides of the figure ;^ 
that is, by the foregoing corol- 
lary, they are equal to all the 
interior angles of the figure, to- 
gether with four right angles •, therefore all the exterior angles 
are equal to four right angles. 

^ * PROP. XXXIII. THEOR. 

. X HE straight lines which join the extremities of 
two equal and parallel straight lines, towards the 
same parts, arc also themselves equal and parallel. 

Let AB, CD be equal and pa- 
rallel straight lines, and joined^ 
towards the same parts by the 
straight lines AC, BD ; AC, BD 
are also equal and parallel. 

JoinBC; and because A B is 

parallel to CD, and BC meets C 1) 

^- them, the alternate angles ABC, BCD are equal*; and because 
AB is equal to CD,and BC common to the two triangles ABC, 
DCB, the two sides AB, BC are equal to the two DC, CB ; 
and the angle ABC is equal to the angle BCD ; therefore the 
*• 1- base AC is cqual'^ to the base BD, and the triangle ABC to the 
triangle BCD, and the other angles to the other angles'", each 
to each, to which the equal sides are opposite ; therefore the 

angle 




» ng 



OF EUCLID. 3i 

angle ACB is equal to the angle CBD ; and becaufe the ^^ 
ftraight line BC meets the two itraight lines AC, BD, and 
makes the alternate angles ACB, CBD equal to one another, 
AC is parallel"^ to BD ; and it was ihewn to be equal to it. " - 
Therefore, ftraight lines, 6cc. Q_ E. D. 



PROP. XXXIV. THEOR. 

A HE opposite sides and anc^les of parallelograms 
are equal to one another, and the diameter bisects 
them, that is, divides them into two equal parts. 

N. B. A parallelogram is a four-sided Jigure, of 
u hich theoppositesidesarc parallel; and the diame- 
ter is the straight line joining tuo of its opposite 
angles. 

Let ABCD be a parallelogram, of which BC is a diame- 
ter ; the oppofice lides and angles of the figure are equal to one 
another ; and the diameter BC bifecls it. 

Becaufe AB is parallel to CD, a 
and BC meets them, the alter- 
nate angles ABC, BCD are equal 

*to one another; and becaufe » ^ \ aoo i 

AC is parallel to BD, and BC \ «y. i. 

meets them, the alternate angles 
ACB, CBD are equal* to one 

another ; wherefore the two triangles ABC, CBD have two 
angles ABC, BCA in one, equal to two angles BCp, CBD 
in the other, each to each, and one fide BC common to the 
two triangles, which is adjacent to their equal angled; there- 
fore their other fides {hall be equal, each to each, and the third 
angle of the one to the third angle of the other,'' viz. the fide'' 
AB to the fide CD, and AC to BD, and the angle BAC equal 
to the angle BDC : And becaufe the angle ABC is equal to 
the angle BCD, and the angle" CBD to the angle ACB, the 
whole angle ABD is equal to the whole angle ACD : And the 
angle BAC has been Ihewn to be equal to the angle BDC ; 
therefore the oppofite fides and angles of parallelograms are 
equal to one another ; alfo, their diameter bifecls thern ; for AB 
being equal to CD, and BC common, the two AB, BC are 
rqual to the two DC, CB, each to eachj and the aneL- ABC 




32 THE ELEMENTS 

^°^^ is equal to the angle BCD ; therefore the triangle ABC is 
'^4.i. equal'= to the triangle BCD, and the diameter BC divides the 
parallelogram ACDB into two equal parts. Q^ E. D. 



SeeN. 



Sec the 2d 
and 3d fi- 
gures, . 



»34, 1. 



1 Ax. 



= 2. or 3, 
Ax. 



<29. 1. 
= 4. 1. 



• 3. Ax, 



PROP., XXXV. THEOR. 

-I Aral LELO GRAMS upon the same base, and be- 
tween the same parallels, are efjual to one another. 

Let the parallelograms ABCD, EBCF be upon the fame 
bafe BC, and between the fame parallels AF, BC ; the paral- 
lelogram ABCD fhali be equal to the parallelogram EBCF. 

it the fides AD, DF of the paral- 
lelograms ABCD, DBCF, oppofite a Ti TT 

to the bafe BC, be terminated in the ~ 

fame point D ; it is plain that each of 
the parallelograms is double* of the 
triangle BDC ; and they are there- 
fore equal to one another. - 

But, if the fides AD, EF, oppofite^ 
to the bafe BC of the parallelograms 

ABCD, EBCF, be not terminated in the fame point ; then, be- 
caufe ABCD is a parallelogram, AD is equal* to BC ; for the 
fame reafon EF is equal to BC j wherefore AD is equal'' to 
EF ; and DE is common ; therefore the whole, or the remain- 
der, AE is equal'^ to the whole, or the remainder DF ; AB alfo 
is equal to DC; and the two EA, AB are therefore equal to 

D F 





the two FD, DC, each to each ; and the exterior angle FDC is 
equal'' to the interior EAB, therefore the bafe EB is equal to 
the bafe FC, and the triangle E A B equal'^ to the triangle FDC : 
take the triangle FDC from the trapezium ABCF, and from 
the fame trapezium take the triangle EAB : the remainders 
therefore are equaF, that is, the parallelogram ABCD is equal 
to the parallelogram EBCF. Therefore parallelograms upon 
the fame bafe, &c. Q^ E. P. 





OF EUCLID. 



PROP. XXXVI. THEOR. 

X ARALLELOGRAMS upon equal bases, and 
between the same parallels, are equal to one another. 

Let ABCD, EFGH be ^ p. ^ 

parallelograms upon equal 

bases BC, FG, and be- 
tween the same parallels 
AH, BG ; the paraHelo- 
gram ABCD is equal to 
EFGH. i^ ,, ^ 

J.vin BE, CH ; and be- '> ^ ^ G 

cause BC is equal to FG, and FG to' EH, BC is equal to *54. i. 
EH : and thev are parallels, and joined towards the same parts 
by t'le straight lines BE, CH : But straight lines which join 
equal and parallel straight lines towards the same parts, are 
themselves equal and parallel ;'' therefore EB, CH, are both "33.1. 
equal and parallel, and EBCH is a parallelogram j and it is 
equal*^ to ABCD, because it is upon the same base BC, and "^5. i, 
between the same parallels ^C, AD : For the like reason, the 
parallelogram EFGH is equal to the same EBCH : There- 
fore also the parallelogram ABCD is equal to EFGH. — 
Wherefore parallelograms, &c. Q. E. D. 

PROP. XXXVn. THEOR. 

I RI ANGLES upon the same base, and between 
the same parallels, are equal to one another. 

Let the triangles ABC, DBC, be upon the same base BC,and 
between the same parallels 17 \ 1> V 

AD, BC: The triangle ABC -' -^^^ — — ^ 

is equal to the triangle DBC. 

Produce AD both ways to 
the points E, F, and through \ / / \ \ / ^ . 

B draw^ BE parallel to CA ; ^ ^^ ^^ ^ *^^' ' 

and through C draw CF pa- 
rallel to BD : Therefore each 
of the figures EBCA, DBCF is a parallelogram ; and EBC A 
is equal^ to DBCF, because they are upon the same .base BC, " 35. 1. 
and. between the same parallels BC, EF ; and the triangle 
ABC is the half of the parallelogram, EBCA, because the 

D diameter 





THE ELEMENTS 

diameter AB bisects <= it ; and the triangle DBC is the half of 
the parallelogram DBCF, because the diameter DC bisects 
it : but the halves of equal things are equal i*^ therefore the 
triangle ABC is equal to the triangle DBC. Wherefore 
triangles, Sec. Q^ E. D. 



PROP. XXXVIII. THEOR. 

J. RIANGLES upon equal bases, and between the 
same parallels, are equal to one another. 

Let the triangles ABC, DEF be upon equal bases BC, EF, 
and between the same parallels BF, AD : The triangle ABC 
Js equal to the triangle DEF. 

Produce AD both ways to the points G, H, and through 
B draw BG parallel * to CA, and through F draw FH paral- 
lel to ED : Then each p a y>. Vi 

of the figures GBCA,f ~^^ -^ ^^ 

DEFH, is a parallel- 
ogram; and they are 

* 36. 1. equal to'= one another, 
because they are upon 

equal bases BC, EF, 

and between the same t5 i.- IL r 

< 34. 1. parallels BF, GH ; and the triangle ABC is the half *^ of the 
parallelogram GBCA, because the diameter AB bisedts it ; and 
the triangle DEF is the half "^ of the parallelogram DEFH, 
because the diameter DF bise£ls it : But the halves of equal 

•^ 7. Ax. things are equal ;'' therefore the triangle ABC is equal to the 
triangle DEF. Wherefore triangles, &c. Q^ E. D. 



•31. 1. 




PROP. XXXIX. THEOR. 



Equ. 



lLQUAL triangles upon the same base, and upon 
the same side of it, are between the same parallels. 

Let the equal triangles ABC, DBC be upon the same base 
BC, and upon the same side of it ; they are between the same 
parallels. 

Join AD ; AD is parallel to BC ; for, if it is not, through 
• 31.-1. the point A draw^ AE parallel to BC, and join EC : The 

X triangle 



I 



OF EUCLID. 

triangle ABC is equal '' to the triangle a 
EBC, because it is upon the same base ^ 
BC, and between the same parallels 
EC, AE : But the triangle ABC is 
equal to the triangle BDC ; therefore 
also the triangle BDC, is equal to the 
triangle EBC, the greater to the less, 
which is impossible : Therefore AE is '^ ^ 

not parallel to BC. In the same manner, it can he demon- 
strated, that no other line but AD is parallel to BC ; AD is 
therefore parallel to it. Wherefore equal triangles upon, &c. 
Q. E. D. 



35 

Book I. 




PROP. XL. THEOR. 



E. 



QIJAL triangles upon equal bases, in the same 
straight line, and towards the same parts, are be- 
tween the same parallels. 

Let the equal triangles ABC, DEF be upon equal bases BC, 
EF, in the same straight ^ D 

line BF, and towards the 
same parts ; they are be- 
tween the same parallels. 

Join AD ; AD is paral- 
lel to BC : For, if it is 
not, through A draw^ AG 
parallel to BF, and join 
GF : The triangle ABC is equal'' to the triangle GEF, be- i 
cause they are upon equal bases BC, EF, and between the 
same f>arallels BF, AG : But the triangle ABC is equal to 
the triangle DEF ; therefore also the triangle DEF is equal 
to the triangle GEF, the greater to the less, which is impossi- 
ble : Therefore AG is not parallel to BF : And in the same 
manner it can be demonstrated that there is no other parallel 
to it but AD, AD is therefore parallel to BF. Wherefore 
equal triangles, kc. Q. E. Dt 




»31. 1. 



33. 1. 



PROP. XLI. THEOR. 

. F a parallelogram and triangle be upon the same 
base, and between the same parallels ; the parallel- 
ogram shall be double of the triangle. 

D2 Let 



36 



Book I. 



•37. 1. 



"34.1, 



THE ELEMENTS 

Let the parallelogram ABCD and the triangle EBC be upon 
the same base BC, and between the same parallels BC, AE ; 
the parallelogram ABCD is double of ^ T\ -p 
the triangle EEC. "^ ' 

Join AC ; then the triangle ABC is 
equal ^ to the triangle EBC, because 
they are upon the same base BC, and 
between the same parallels BC, AE. 
But the parallelogram ABCD is 
double'' of the triangle ABC, because 
the diameter AC divides it into two ; " C 

equal parts ; wherefore ABCD is also double of the triangle 
EBC. > Therefore, if a parallelogram, &c. Q_. E. D. 




PROP. XLIL PROB. 



» 10. 1. 
•^31.1. 



"38. 1. 



«4i. 1. 



JL O describe a parallelogram that shall be equal 
to a given triangle, and have one of its angles equal 
to a given rectilineal angle. 

Let ABC be the given triangle, and D the given reftilineal 
angle. It is required to describe a parallelogram that shall be 
equal to the given triangle ABC, and have one of its angles 
equal to D. 

Bisect^ BC in E, join AE, and at the point E in the straight 
line EC make'' the angle CEF equal to D ; and through A 
draw*^ AG parallel to EC, and * v r* 

through C draw CG"= parallel to '^~* ' 

EF : Therefore FECG is a 
parallelogram: And because BE 
is equiil to EC, the triangle ABE 
is likewise equal'' to the triangle 
AEC, since they are upon equal 
bases BE, EC, and between the /^ 
same parallels BC, AG ; there- H 
fore the triangle ABC is double of the triangle AEC. And 
the parallelogram FECG is likewise double*' of the triangle 
AEC, because it is upon the same base, and between the san>e 
parallels : Therefore the parallelogram FECG is equal to the 
triangle ABC, and it has one of its angles CEF equal to the 
give-a angle D j wherefore there has been described a parallel^. 

ogram 





OF EUCLID. 37 

©gram FECG equal to a given triangle ABC, having one of ^<">^ I- 
its angles CEF equal to the given angle D. Which was to ^"^"^^^^ 
be done. 

PROP. XLIII. THEOR. 

A HE complements of the parallelograms, which 
are about the diameter of any parallelogram, ^re 
equal to one another. 

LwC ABCD be a parallelogram, of which the diameter is 
AC, and LH, FG, and parai- ^.^ J^ 
lelograms ao lut AC, that is 
through ivhLh AC passes.^ and 
BK, KD, the other parallelo- 
grams wh ch make up the whole 
figure ABCD, which are there- 
fore called the complements : 

The complement BK is equal ^ 

to the complement KD. Vi G ' C 

Because ABCD is a parallelogram, and AC ks diameter, , 
the triangle ABC is equal' to the triangle ADC : And, because * " ^' 
EKHA IS a parallelogiam, the diametec of which is AK, the 
triangle AEK is equal to the triangle AHK : By the same 
reason, the triangle KGC is equal to me triangle KFC: Then, 
because the trianglr AEK is equal to the triangle AHK, and 
the triangle KGC to KFC ; the triangle AEK, together with 
the triangle KGC is equal to the triangle.AHK together with 
the triangle KFC : But, the whole triangle ABC is equal to 
the whole ADC i ihereiore the remaining complement BK is 
equal to the remaining complement KD. Wherefore the 
complements, &c. Q^ E. D. 

PROP. XLIV. PROB. 

X O a given stmight line to apply a parallelogram, 
which shall be equal to a given tnangle, and have 
one of its angles equal to a given rectilineal angle. 

Let AB be the given straight line, and C the given triscn- 
gle, and D the given rectilineal angle. It is required toap- 
ply to the straight line AB a parallelogram equal to the tri- 
angle C, and having an angle equal to D. 

D3 Make 



38 



THE ELEMENTS 



»>31. 1. 



29. 1. 



« 12. Ax 



^43. 1. 
n5. 1. 




, Make* the 
parallelogram 
BEFG equal 
to the triangle 

C, and hav- 
ing the angle 
EBG equal 
to the angle 

D, so that BE 
'he in the same 

straight line with AB, and produce FG to H ; and through A 
draw'' AH parallel to BG or EF, and join HB. T^hen be- 
' cause the straight line HF falls upon the parallels AH, EF, 
the angles AHF, HFE, are together equa^ to two right an- 
gles ; wherefore the angles BHF, HFE are less than two right 
angles : But straight lines which with another straight line 
tnake the interior angles upon the same side less than two right 
angles, do meet'' if produced far enough : Therefore HB, FE 
shall meet if produced ; let them meet in K, and through K 
draw KL, parallel to EA or FH, and produce HA, GB to the 
points L, M : Then HLKF is a parallelogram, of which the 
diameter is HK, and AG, ME are the parallelograms about 
HK ; and LB, BF are the complements: therefore LB is equal^ 
to BF ; But BF is equal to the triangle C ; wherefore LB is 
equal to the triangle C ; and because the angle GBE is equal*^ 
to the angle ABM, and likewise to the angle D ; the angle 
ABM is equal to the angle D : Therefore the parallelogram 
LB is applied to the straight line AB, is equal to the triangle 
C, and has the angle ABM equal to the angle D : Which, 
was to be. done. 



SeeN, 



»42. 1. 



'44.1. 



PROP. XLV. PROB. 

L O describe a parallelogram equal to a given rec- 
tilineal figure, and having an angle equal to a given 
rectilineal angle. 

Let ABCD be the given redilineal figure, and E the given 
rectilineal anpjle. It is required to describe a parallelogram 
equal to ABCD, and having an angle equal to E. 

Join DB, and describe'' the parallelogram FH equal to the 
triangle ADB, and having the angle HKF equal to the angle 
E i and to' the straight line GH apply^ the parallelogram GM 

equai 



OF EUCLID. 



39 




2i>. 1. 



equal to the triangle DBC, having the angle GHM equal to BookI. 
the angle E ; and because the angle E is equal to each of the ^'^^■''^^ 
angles FKH, GHM, the angle EKH is equal to GHM : add 
to each of these the angle KHG ; therefore the angles FKH, 

KHG are equal A 1) p^ G L, 

to the angles 
KHG, GHM; 
butFKH,KHG 
are equal '^ to 
two right angles; 
Therefore also 
KHG, GHM, 
are equal to two 
right angles; and 
because at the point H in the straight line GH, the two 
straight lines KH, HM, upon the opposite sides of it make 
the adjacent angles equal to two right angles, KH is in the 
same straight line** with HM ; and because the straight line " U. i. 
HG meets the parallels KM, FG, the alternate angles MHG, 
HGF are equal := Add to each of these the angle HGL : 
Therefore ihe angles MHG, HGL, are equal to the angles 
HGF, HGL : But the angles MHG, HGL, are equal " to 
two right angles ; wherefore also the angles HGF, HGL are 
equal to two right angles, and FG is therefore in the same 
straight line with GL ; and because KF is parallel to HG, and 
HG to ML ; KF is parallel to ML ; and KM, FL are pa-^so i 
rallek ; wherefore KFLM is a parallelogram ; and because the 
triangle ABD is equal to the parallelogram HF, and the tri- 
angle DBC to the parallelogram GM; the whole reililineal 
figure ABCD is equal to the whole parallelogram KFLM ; 
therefore the parallelogram KFLM has been described equal 
to the given reftilineal figure ABCD, having the angle FKM 
equal to the given angle E. Which was to be done. 

Cgr. From this it is manifest how to a given straight line 
to apply a parallelogram, which shall have an angle equal to a 
given re£Vilineal angle, and shall be equal to a given rectilineal 
figure, viz. by applying'' to the given straight line a parallel- 1> 4^. :. 
ogram equal to the first triangle ABD, and having an angle 
«qual to the given angle. 



04 



40 THE ELEMEI^TS 

Book I. 

PROP. XLVI. PROB. 



X O describe a square upon a given straight line. 

Let AB be the given straight line; It is required to de- 
scribe a square upon AB. 
Ml. 1. Frdm the point A draw^ AC at right angles to AB ; and 

'3. ], make^ AD equal to AB, and through the point D draw DE 
«3i. 1, parallel to AB, and through B draw BE parallel to AD ; 
<i34 1 therefore ADEB is a parallelogram : whence AB is equal'' to 
DE, and AD to BE : But B A is equal (^ 
to AD; therefore the four straight lines 
BA, AD, DE, EB are equal to one 
-another, and the parallelogram ADEB -t) 
is equilateral, likewise all its angles are 
right angles ; because the straight line 
AD meeting the parallels AB, DE, 
^29.1. the angles BAD, ADE are equal-^ to 
two right angles ; but BAD is a right 
angle ; therefore also ADE is a right 

angle ; but the opposite angles of paral- Al B 

lelograms are equal ;•* therefore each of the opposite angles 
ABE, BED is a -right angle*; wherefore the figure ADEB is 
rectangular, and It has been demonstrated that it is equila- 
teral; it is therefore a square, and it is described upon- the 
given straight line AB ; Which was to be done. 

Cor. Hence every parallelogram that has one right angle 
has all its angles right angles. 



PROP. XLVII. THEOR. 

J. NT any right-angled triangle, the square which is 
described upon the side subtending the right angle, 
is equal to the squares described upon the sides 
which contain the right angle; 

Let ABC be a right-angled triangle having 'the right angle 
BAC ; the square described upon the side BC is equal to the 
squares described upon BA, AC. 
''^- 1- On BC describe'' the square BDEC, and on BA, AC the 

square? 





«fu;». 



OF EUCLID. 

squares GB, HC ; and through A draw *> AL parallel to BD, 
or CE, and join AD, FC ; then, because each of the angles 
BAG, BAG is a right angles 
the two straight lines AC, 
AG, upon the opposite side* 
of A B, make witn ii ac the 
point A the adjacent angles 
' equal to two right angles ; 
therefore CA is in the sarrie 
straight line ^ with AG; for 
the same reason, AB and AH 
are in the same straight line ; 
and because the angle DBC is 
equal to the ar^gle FBA, each 
ot them being a right annle, 

add to each the angle ABC, 

and the whole angle DBA is equal ^ to the whole FBC ; and ' 2 Ak. 
because the two sides AB, BD are equal to the two FB, BC, 
each to each, and the angle DBA equal to the angle FBCj 
therefore the base AD is equal ^ to the base FC, and the trian- f 4. i. 
gle ABD to the triangle FBC : Now the parallelogram BL is 
double E of the triangL ABD, because they are upon the saoie s 41. i. 
base BD, and between the same parallels, BD, AL ; and the 
square G3 is double of the triangle FBC, because these also 
are upon the same base FB, and between the same parallels 
FB, GC. But the doubles of equals are equal •" to one ano- » 6. .\jt 
ther ; Therefore the parallelogram BL is equai to the square 
GB: And, in the same manner, by J9ining AE, BK, it is de- 
monstrated, that the parallelogram CL is equal to the square 
HC ! Therefore the whole square BDEC is equal to the two 
squares GB, HC ; and the square BDEC is described upon 
the straight line BC, and the squares GB, HC upon BA, AC : 
Wherefore the square upon the side BC is equal to the squares 
upon the sides BA, AC Therefore, in any right-angled tri- 
angle, &c. Q, E. D. 



PROP. XLVIIL THEOR. 



If the square described upon one of the sides of a 
triangle, be equal to the squares describedupon the 
other two sides of it ; the aujrle contained bv these 



two sides is a riiiht an«:le. 



If 



42 

Book I. 



• JI. 1. 



47.1. 



^6. 1. 



THE ELEMENTS, &c. 

If the square described upon BC, one of the sides of the tri- 
angle ABC, be equal to the squares upon the other sides BA, 
AC, the angle BAC is a right angle. 

From the point A draw* AD at right angles to AC, and 
make AD equal to BA, and join DC : Then, because DA is 
equal to AB, the square of DA is equal to 
the squares of AB : To each of these add 
the square of AC ; therefore the squares 
of Da, AC are equal to the squares of 
BA, AC : But the square of DC is cqual^ 
to the squares of DA, AC, because DAC 
is a right angle ; and the square of BC, 
by hypothesis, is equal to the squares of 
BA, Ac ; therefore the square of DC is ^^ 
equal to the square of BC ; and therefore o 
also the side DC is equal to the side BC. And because the 
side DA is equal to AB, and AC common to the two triangles 
DAC, BAC, the two DA, AC are equal to the two BA, AC j 
and the base DC is equal to the base BC ; therefore the angle 
DAC is equal <= to the angle BAC ; but DAC is a right an-^ 
gle; therefore also BAC is a right angle. Therefore, if the 
S(juare, &c. Q. E. D, 




THE 



Book. 11. 



ELEMENTS 



EUCLID, 



BOOK. II. 



DEFINITIONS. 

I. 

ijjVERY right angled parallelogram is said to be 
contained by any two of the straight lines which 
contain one of the right angles. 

II. 
In every parallelogram, any of the parallelogranis about a diju. 
meter, together with the two 
complements, is called a Gno- 
mon. ' Thus the parallelo- 
' gram HG, together with the 

* complements AF, FC,is the 

* gnomon, which is more 

* briefly expressed bvthe let- 
< ters AGK, or EHC, which 

* are at the opposite angles of ^ 

* the parallelograms which make the gnomon.' 

PROP. I. THEOR. 

If there be two straight lines, one of which is di- 
vided into any number of parts ; the rectangle con- 
tained by the two straight lines, is equal to the rec- 
tangles contained by the undivided line, and the 
several parts of the divided line. 





n. 1. 



*3. J. 
'31. 1. 



54. ]. 



THE ELEMENTS 

Let A and BC be two straight lines ; and let BC be divided 
into any partsin the points D, E j the rectangle contained by 



B 




the straight lines A, BC is equal 
to the rectangle contained by 
A, BD, together with that con- 
tained by A, DE, and that con- 
tained by A, EC. 

From the point B draw^ BF ; Q 
at right angles to BC,ancl make 
BG equal '" to A ; and through 
G draw "^ GH parallel to BC ; 
and through D, E, C, draw ^ DK, EL, CH parallel to BG; 
then the rectangle BH is equal to the rectangles BK, DL, 
EH ; and BH is contained by A, BC, for it is contained by 
GB, bC, and GB is equal to A ; and BK is contained 
by A, BD, for it is contained by GB, BD, of which GB 
is equal to A ; and DL is contained by A, DE, because 
DK, that is ^ BG, is equal to A ; and in like manner the rect- 
angle EH is contained by A, EC : Therefore the rectangle 
contained by A, BC, is equal to the several rectangles con- 
tained by A, BD, and by A, DE ; aiid also by A, EC. 
Wherefore, if there be two straight lines, Sic. Q^ E, D. 



•4i>. 1, 



. PROP. IL THEOR. 

XT a straight line be divided into any two parts, 
the rectangles contained by the whole and each of 
the parts, are together equal to the square of the 
whole line. 

Let the straight line AB be divided 
into any two parts in the point C; the 
rectangle contained by AB, BC ; toge- 
ther with the rectangle* AB, AC, shall 
be equal to the square of AB. 

Upon AB ^describe » the sqbare 
ADEB, and' through C draw ^ CF, 
parallel to AD or BE ; then AE is equal 
to the rectangles AF, CE ; and AE is _ ^ ^ 

the square of AB j and AF is the rectangle contained'by BA, 
AC ; for it is contained by DA, AC of which AD is equal 

B. To 




*N 



. B. To avoid repratini; the word contained Xoo freqoentlv, tl.c rectangle ' 
by two straight lines AB. AC is sometimes simply called t!ie rectangle AB, 



con- 
AC. 

to 



OF EUCLID. 



45 



to AB J and CE is contained by AB, BC, for BE is equal to ^^^^ 
AB ; therefore the rectangle contained by AB, AQ together ''^"'''''^ 
with the rectangle AB, BC, is equal to the square of AB. If 
therefore a straight line, ice. Q: E. D. 



PROP. III. THEOR. 

F a straight line be divided into any two parts, 
the rectangle contained by the whole and one of the 
parts, is equal to the rec tingle contained by the two 
parts, together with the square of the aforesaid part. 

Let the straight line AB be divided into two parts in the 
point C ; the rectangle AB, BC, is equal to the recungle AC, 
CB, together with the square of BC. 

Upon BC describe * the square A C B * ^^- ^ 

CD£B, and produce ED to F, and 

through A draw^'AF parallel to i>5i_ j^ 

CD or BE ; then the rectangle AE 

is equal to the rectangles AD, CE i 

ajwJ AE is the rectangle contained 

by AB, BC, for it is contained by, 

AB, BE, of which BE is equal to 

BC ; and AD is contained by AC, V 

CB, for CD is equal to BC ; and DB is the square of BC ; 

therefore the rectangle AB, BC, is equal to the rectangle AC, 

CB, together with the square of BC. If therefore a straight 

line, &c. Q. E. D. 




PROP. IV. THEOR. 

1 F a straight line be divided into any two parts, the 
square of the whole line is equal to the squares of 
tlie two parts, together with twice the rectangle 
contained by the parts. 

Let the straight line AB be divided into any two parts in 
C ; the square of AB is equal to the squares of AC, CB, and 
to twice the rectangle contained by AC, CB. 

Upon 



■/ 

^ 



46 

Book II. 

»46. 1. 
«>31. 1. 



' 29. 1. 
«»5. 1. 



e6. 1. 
<"34.. 1. 



e43. 1. 




THE ELEMENTS 

Upon AB describe =» the square ADEB, and join BD, ancJ 
through C draw*- CGF parallel to AD or BE, and through G 
draw HK parallel to AB or DE: And because CF is parallel 
to AD, and BD falls upon them, the exterior angle BGC is 
equal"^ to the interior and opposite angle ADB ; but ADB is 
equal <^ to the angle ABD, because BA is equal to AD, being 
sides of a square ; wherefore the an- 
gle CGB is equal to the angle GBC ; 
and therefore the side BC is equal ^ 
to the side CG: But CB is equal* also 
to GK, and CG to BK ; wherefore 
the figure CGKB is equilateral : It is 
likewise rectangular ; for CG is pa- 
rallel to BK, and CB meets them j 
the angles KBC, GCB are therefore 
equal to two right angles ; and KBC is a right angle ; where- 
fore GCB is a right angle ; and therefore also the angles ^ 
CGK, GKB opposite to these, are right angles, and CGKB 
is rectangular; but it is also equilateral, as was demonstrated ; 
wherefore it is a square, and it is upon the side CB : For the 
same reason HF also is a square, and it is upon the side HG, 
which is equal to AC : Therefore HF, CK are the squares 
of AC, CB ; and because the complement AG is equal e 
to the complement GE, and that AG is the rectangle con- 
tained by AC, CB,. for GC is equal to CB ; therefore GE is 
also equal to the irectangle AC, CB ; wherefore AG, GE are 
equal to twice the rectangle AC, CB : And HF, CK are the 
squares of AC, CB ; wherefore the four figures HF, CK, 
AG, GE are equal to the squares of AC, CB, and to twice 
the rectangle AC, CB : But HF, CK, AG, GE make up 
the whole figure ADEB, which is the square of AB : There- 
fore the square of AB is ^qual to the squares of AC, CB, and 
twice the rectangle AC, CB. Wherefore if a straight line, 
&c. Q, E. D. 

CoR. From the demonstration, it is manifest, that the pa- 
rallelograms about the diameter of a square arc likewise 
squares. 



OF EUCLID. 



47 

Book II. 



PROP. V. THEOR. 

JtF a straight line be divided into tw^o equal parts, 
and also into two unequal parts, the rectangle con- 
tained by the unequal parts, together with the 
square of the line betAveen the points of section, is 
-equal to the square of half the line. 

Let the straight line AB be divided into two equal parts in 
the point C, and into two unequal parts at the point D; the 
redangle AD, DB, together with the square of CD, is equal 
to the square of CB. 

Upon CB describe" the square CEFB, join BE, and through * -i^- ^• 
Ddraw" DHG parallel to CE or BF ; and through H draV ■> 31. i. 
KLM parallel to CB or EF ; and also through A draw AK 
parallel to CL or BM : And because the complement CH is 
equal'-" to the complement HF, to each of these add DM; there- c 43. 1. 
fore the whole CM is equal J^ 
to the whole DF ; but CM '" 
Is equaH to AL, because 
AC is equal to CB ; there- }^ 
fore also AL is equal to DF. 
To each of these add CH, 

and the whole AH is equal 

to DF and CH : But AH E G ¥ 

is the redangle contained by AD, DB, for DH is equal = tO'Cor. 4. 2. 

DB i and DF together with CH is the gnomon CMG ;' 

therefore the gnomon CMG is equal to the rectangle AD, 

DB : To each of these add LG, which is equal= to the square 

of CD ; therefore the gnomon CMG, together with LG, 

is equal to the re6tangle AD, DB, together with the square 

of CD : But the gnomon CMG and LG make up the whole 

figure CEFB, wiiich is the square of CB : Therefore the 

retlangle AD, DB, together with the square of CD, is equal 

to the square of CB. Wherefore^' if a straight line, &c. 

Q, E. D. 

From this proposition it is manifest, that the difference of 
the squares of two unequal lines AC, CD, is equal to the 
re<5langle contarned by their sum and difference. 




36. 1. 



^ 



Book U. 



TriE ELEMENTS 



PROP. VI. THEOR. 



•46. 1. 
•31. L 



«45. 1 



J.F a straight line be bisected, and produced to any 
point; the rectangle contained by the whole line 
thus produced, and the part of it produced, toge- 
ther with tiie square of half the line bisected, is equal 
to the square of the straight line which is made up 
of the half and tlie part producetl. 

Let the straight line AB be bisected in C, and produced to 
the point D ; the reSangle AD, DB, together with the square 
of CB^ is equal to the 'square ot CD. 

UponCDdescribeHhe square CEFD,join DE,and through 
B draw'' BHG parallelto, CE or DF, and through H draw 
KLM parallel to AD or EF, and also through A draw AK pa- 
rallel to CL or DM; and be- ^ (3 ^ J) 
cause AC is equal to CB, 
the re6langle AL is equal^ to 
CH ; but CH is equaF toy^ 
HF : therefore also AL is 
equal to HF: To each of these 
add CM; therefore the whole 
AM is equal to the gnomon 
CMG: And AM is the 
< Cor. 4.2. reilangle contained by AD, DB, for DM is equa^ to DB : 
Therefore the gnomon, CMG is equal to the rc6langle AD, 
DB : Add to each of these I/G, which is equal to the square 
of CB, therefore the reftangle AD, DB, together with the^ 
square of C3, is equal to the gnomon CMG, and the figure LG; 
But the gnomon CMG and LG make up the whole figure 
CEFD, which is the square of CD ; therefore the redlangie 
AD, DB, together with the square of CB, is equal to the 
square oi CD. W herefore, if a straight line, Sec. Q^ E. D. 

PROP. VIL THEOR. 

If a straight line he divided into any two parts, 
the squares of the whole line, and -of one of the 
parts, arc equal to twice the rectangle contained by 
the whole and that part, together with the square 
of the other part. 

Let the straight line AB be divided into any two parts in 

the 




OF EUCLID. 



49 



the point C; the squares of AB, EC are equal to twice the Book.II. 
reftangle AB, BC, together with the square of AC. ^.-"v^ 

' Upon AB describe * the square ADtB, and construft t^e* ^o. i. 
figure as in the preceding propositions; and because AG is 
equal'' to GE, add to each of them CK ; the whole AK is* 40. i. 
therefore, equal to the whole CE j 
therefore AK, CE, are double of 
AK : But AK, CE, are the gnomon A 
AKF, together with. the square CK ; 
therefore the gnomon AKF, toge- ^ 
ther with the square CK, is double 
of AK : But twice the rectangle AB, 
BC is double of AK, for BK is 
equa^ to BC : Therefore the gno- 
mon AKF, together with the square j^"^ 
CK, is equal to twice the re^angle 
AB, EC : To each of these equals 

add HF, which is equal to the square of AC ; therefore the 
gnomon AKF, together with the squares CK, HF, is equal to 
twice the re<5langle AB, BC, and the square of AC : but the 
gnomon AKF, together with the squares CK, HF, make up 
tiie whole figure ADEB and CK, which are the squares of AB 
and BC : therefore the squares of AB and BC are equal to 
twice the rectangle AB, BC, together with the square of AC. 
Wherefore, if a straight line, &c. Q. E. D. 




« Cor. 4. C. 



PROP. VIII. THEOR. 

J.F a straight line be divided into any two parts, 
four times the rectangle contained by the whole line, 
and one of the parts, together with the square of 
the other parr, is equal to the square of the straight 
line, which is made up of the whole and that part. 

Let the straight line AB be divided into any two parts in 
the point C-, four times the redangle AB, BC, together with 
the square of AC, is equal to the square of the straight line 
made up of AB and BC together. 

Produce AB to D, so that BD be equal to CB, and upon 
AD describe the square AEFD ; and construct two figures 
such as in the preceding. Because CB is equal to BD, and 
that CB is equal* to GK, and BD to KN ; therefore GK is »34. 

£ equal 



50 

Book IT. 



THE ELEMENTS 




•43.1. 



X 



equal to KN : For the same reason, PR is equal to RO ; and 
because CB is equal to BD, and GK to KN, the redlangle 
CK is equal'' to BN, and GR to RN : but CK is equal^ to 
RN, because they are the complements of the parallelogram 
CO ; therefore also BN is equal to GR ; and the four redit- 
angles BN, CK, GR, RN are therefore equal to one another, 
and so are quadruple of ont of them CK : Again, because CB 
is equal to BD, and that BD is 
^ Cor. 4. 2. equaH to BK, that is, to CG , 

and CB equal to GK, that ^ is, to a 
GP ; therefore CG is equal to 
GP : And because CG is equal to T\/r 
' GP, and PR to RO, the reftangla ^ ^ 
AG is equal to MP, and PL to 
RF: But MP is equal ^ t» PL, 
because they are the complements 
of the parallelogram M L ; where- 
fore AG is equal also to RF : 
Therefore the four redlangles AG, 
MP, PL, RF are equal to one 
another, and so are quadruple of one 

of them AG. And it was demonstrated that the four CK 
BN, GR, and RN are quadruple of CK. Therefore the 
eight rectangles which contain the gnomon AOH, are quad- 
ruple of AK ; .and because AK is the rectangle contained 
by AB, BC, for BK is equal to BC, four times the redangle 
AB, BC is quadruple of AK : But the gnomon AOH was 
demonstrated to be quadruple of AK : therefore four times 
the rectangle AB, BC, is equal to the gnomon AOH. To 
each of these add XH, which is equaK to the square of AC : 
Therefore four times the rectangle AB, BC together with the 
squareof AC, is equal to the gnomon AOH and the square XH: 
But the gnomon AOH and XH make up the figure AEFD, 
which is the square of AD : Therefore four times the redtangle 
AB, BC, together with the square of AC, is equal to the 
square of AD, that is, of AB and BC added together in one 
straight line. Wherefore, if a straight line, &c. Q;. E. D. 




^ Cor. 4.2. 



OF EUCLID. 5, 

Book II. 

. PROP. IX. THEOR. 

J.F a straight line be divided into two ecjual, and 
also into two unequal parts; the squares of the tv/o 
unequal parts are together double of the square of 
half the line, and of the square of the line between 
the points of section. 

Let the straight line Ah be divided at the point C into two 
equal, and at D into two unequal parts : The squares of AD, 
DB are together double of the squares of AC, CD. 

From the point C draw* CE at right angles to AB, and Mi. ji. 
make it equal to AC or CB,and join E A, EB ; through D draw 
"DF parallel to CE, and through F draw FG parallel to AB ; " si. }. 
and join x\F ; Then, because AC is equal to CE, the angle 
EAC is equal<= to the angle A EC ; and because the angle "^ 5.1. 
ACE is a right angle, the two others AEC, EAC together 
make one right arigle"* ; and they are equal to one another •,"22.1. 
each of them therefore is half of 
a right angle. For the same 
reason each of the angles CEB, 
EBC is half a right angle j and 
therefore the whole AEB is a 
right angle : And because the an- 
gle GEF is half a right angle, 
and EGF a right angle, for it is 

cquaP to the interior and opposite angle ECB, the remaining eoo. 1, 
angle EFG is half a right angle ; therefore the angle GEF is 
equal to the angle EFG, and the side EG equaF to thcfg, j, 
side GF; Again, because the angle at B is half a right 
angle, and FDB a right angle, for it is equal ' to the in- 
terior and opposite angle ECB, the remaining angle BFD 
is half a right angle ; therefore the angle at B is equal to the 
angle BFD, and the side DF to ^ the side DB : And because 
AC is equal to CE, the square of AC is equal to the square 
of CE i therefore the squares of AC, CE, are double of the 
square of AC : But the square of EA is equals to the squares evT. 1. 
of AC, CE, because ACE is a right angle; therefore the 
square of EA is double of the square of AC : Again, be- 
cause EG is equal to GF, the square of EG is equal to the 
square of GF ; therefore the squares of EG, GF are double of 

E 2 the 



J 




\ 


A ( 


> 


D B 



47. 1. 



52 THE ELEMENTS 

Book u. the squarc of GF i but the squire of EF is equal to the squares 
^"^'"''''^^ of EG, GF ; therefore the square of EF is double of the square 
* 34. 1. Qp . 2j^(j Qp js Qqualh to CD ; therefore the square of EF is 
double of the squarc of CD : But thesguare of AE is likewise 
double of the square of AC ; therefore the squares of AE, EF 
are double of the squares of AC, CD : And the square of AF is 
equal' to the squares of AE, EF, because AEF is a right angle j 
therefore the square of AF is double of the squares of AC, 
CD : But the squares of AD, DF, are equal to the square of 
AF, because the angle ADF is a right angle j therefore the 
squares of AD, DF are double of the squares of AC, CD : And 
DK is equal toDB; therefore the squares of AD, DB arc 
double of the squares of AC, CD. If therefore a straight line. 
Sec. Q.E. D. 



PROP. X. THEOR. 

IF a straight line be bisected, and produced to 
any point, the square of the whole line thus pro- 
duced, and the square of the part of it produced, 
are together double of the square of half the line 
bisected, and of the square of the hne made up of 
the half and the part produced. 

Let the straight line AB be biseded in C and produced to 
the point D ; the squares of AD, DB are double of the squares 
■ of AC, CD. 
* n. 1. From the point C draw* CE at right angles to A, B : And 
make it equal to AC or CB, and join AE, EB ; through E draw 
*3l. 1. bgp parallel to AB, and through D draw DF parallel to CE: 
And because thestraight line EFmeets the parallelsEC,FD, the 
. 29 J angles CEF, EFD are equa^to two right anglts; and therefore 
theanglesBEF,EFDarelessthantwor!ghtangles; butstraight 
lines which with another straight line make the interior angles 
12. Ax. upon the same side less than two right angles, do meet"* if pro- 
duced far enough : Therefore EB, FD shall meet, if produced 
^ towards B,D: Let them meet in G, and join AG: Then, because 

AC is equal to CE, the angle CEA is equal* to the angle 
EAC ; and the angle ACE is a right angle ; therefore each of 
'3i- i- the angles CEA, EAC is half a right angle'. For the same 

reason, 



OF EUCLID, 



53 




«6. 1, 



t34. 1, 



reason, each of the angles CEB, EEC is half a right angle; ^°°^ 
therefore AEB is a right angle : And because EBC is half a ^^"^'^^ 
right angle, DBG is also^ half a right angle, for they are^ ver- '^^s, i. 
tically qppositc i but BDG is a right angle, because u is equalise. 1. 
to the alternate angle DCE j therefore the remaining angle 
DGB is half a right angle, and is therefore equal to thd angle 
DBGj wherefore also the side BD is equal? to the side DG 
Again, because EGF is 

half a right angle, and that fc^ £ 

. the aagle at F is a right an- 
gle, because ii is equal"* 
to the oD-)oS4teangleECD, 
jthc -emaining angle FEG 
Ss :ulf a V.ght angle, and \4 
ej ial to tie angle EGF ; 
v- efsfore also the side 
CF is equals to the side 

F£. And because EC is equal to CA, the square of EC is 
equal to the square of CA ; therefore the squares of EC, CA 
are Jouble of the square of CA : But the square of EA is 
equal' to the squares of EC, C A ; therefore the square of EA * ^^ 
is double of the square of AC : Again, because GF is equal 
to FE, the square of GF is equal R) the square of FE ; and 
therefore the squares of GF, FE are double of the square of 
EF ; But the square of EG is equal- to the squares of GF, FE; 
therefore the square of EG is double of the square of EF : 
And EF is equal to CD ; wherefore the square of EG is dou- 
ble of the square of CD . But it was demonstrated, that the 
square of EA is double of the square of AC ; therefore the 
squares of AE, EG, are double of the squares of AC, CD ; 
And the square of AG is equaU to the squares of AE, EG ; 
therefore the square of AG is double of the squares of AC, 
CD : But the squares of AD, GD, are equal' to the square of 
AG i therefore the squares of AD, DG are double of the 
squares of AC, CD : But DG is equal to DB ; therefore the 
squares of AD, DB are double of the squares of AC, CD. 
Wherefore, if a straight line, Sec. Q^ E. D. 

E3 



S4 

Book H. 



' THE ELEMENTS 



PROP. XI. PROB. 



■»46. 1. 
»'10. 1. 
*3. 1, 



" (5. .2. 



*47. 1. 



X O divide a given straight line into two parts, so 
that the rectangle contained by the whole, and one 
of the parts, shall be equal to the square of the other 
part. 

Let AB be the given straight line ; it is required to divide 
it into two parts, so that the reftangle contained by the 
whole, and one of the parts, shall be equal to the square of 
the other part. 

Upon AB describe^* the square ABDC ; bisect*" AC in E, 
and join BE; produce CA to F, and make ""EF equal to EB, 
and upon AF describe* the square of FGHA ; AB is divided in 
H, so that the redtangle AB, BH, is equal to the square of AH. 

Produce GH toK ; because the straight line AC is biseded 
in E, and produced to the point F, the redlangle, CF, FA, to- 
gether with the square of AE, is equal '^ to the square of EF: 
But EF is equal to EB j therefore the redangle CF, FA, to- 
gether with the square of AE, is equal to the square cf EB : 
and the squares of BA, AE,are equal*^ 
to the square of EB, because the an- 
gle EAB is a right angle ; therefore the 
reftangle CF, FA, together with the 
square of AE, is equal to the squares 
of BA, AE : Take away the square of 
AE, which is common to both, there- 
fore the remaining redlangle CF, FA, 
is equal to the square of AB j and the 
figure FK is the re£langle contained 
by CF, FA, for AF is equal to FG ; 
arid AD is the square of AB ; there- 
fore FK is equal to AD : Take away 
the common part AK, and the remain- 
der FH is equal to the remainder HD : 
And HD is the redlanele contained by AB, BH, for AB is 
equal to BD j and FH is the square of AH. Therefore the 
redangle AB, BH is equal to the square of AH : Wherefore, 
the straight line AB is divided in H, so that the redlangle AB, 
BH, is equal to the square of AH. Which was tQ be done. 




OF EUCLID. 55 

Book II. 

PROP. XII. THEOR. 

IN obtuse angled triangles, if a perpendicular be 
drawn from any of the acute angles to the opposite 
side produced, the square of the side subtending the 
obtuse angle is greater than the squares of the sides 
containing the obtuse angle, by twice the rectangle, 
contained by the side upon which, when produced, 
the perpendicular falls, and the straight line inter- 
cepted without the triangle between the perpendi- 
cular and the obtuse angle. 

Let ABC be an obtuse angled triangle, having the obtuse 
angle ACB, and from the point A let AD be drawn* perpen- » 12. i. 
dicular to BC produced : The square of AB is greater thJui 
the squares of AC, CB, by twice the rectangle BC, CD. 

Because the straight line BD is divided into two parts in the 
point C, the square of BD is equal 

'^o the squares of BC, CD, and A '*'2- 

twice the reftangle BC, CD : To 
each of these equals add the square 
of. DA; and the squares of BD, DA, 
are equal to the squares of BC, CD, 
DA, and twice the rectangle BC, 
CD : But the square of BA is equal 

•^to the squares of BD, DA, be- ^y^ / j c 47. 1, 

cause the angle at D is a right B C 

angle j and the square of CA is 

equal= to the squares of CD, DA : Therefore the square of 
BA is equal to the squares of BC, CA, and twice the rectan- 
gle BC, CD ; that is, the square of BA is greater than the 
squares of BC, CA, by twice the redangle BC, CD. There- 
fore, in obtuse angled triangles, &c. Q. E. D. 

E 4. 




THE EL.EM ENTS 



Book II. 



SeeN, 



12. 1. 



"7. 2. 



47. 1. 



* 16. 1. 



1'2. 2. 



PROP. XIII. THEOR. 

1 N every triangle, the square of the side subtend- 
ing any of the acute angles, is less than the squares 
of the sides containing that angle, by twice the 
rectangle contained by either of these sides, and 
the stra:ight lipe intercepted between the perpendi- 
cular let fall upon it from the opposite angle, and 
the acute angle; 

Let ABC be any triangle, and the angle at B one of its 
acute angles, and upon BC, one of the sides containing it, let 
fall the perpendicular* AD from the opposite angle : The 
square of AC, opposite to the angle B, is less than the squares 
of CB, BA, by twice the reaangle CB, BD. ' 

First, Let AD fail within the triangle ABC ; and because 
the straight line CB is divided into 
two parts in the point D, the 
squares of CB, BD are equal** to 
twice the redtangle contained by 
CB, BD, and the square of DC : 
To each of these equals add the 
square of AD ; therefore the squares 
ot CB, BD, DA, are equal to twice 
the redtangle CB, BD ; and the 
squares ot AD, DC : But the 
squares of AB is equal= to the 

square BD, DA, because the angle BDA is a right angle ; 
and the square of AC is equal to the squares of AD, DC : 
'rherefoi:e the squares of CB, BA are equal tp the square of 
AC, and twice the recStangle CB, BD, that is, the square of 
AC alone is less than the squares of CB, BA \fy twice the 
rcdangle CB, BD. 

Secondly, Let AD fall without 
the triangle ABC : Then, because 
the angle at D is a right angle, 
the angle ACB is greater "^ than 
a right angle; and therefore the 
square of AB is equal *= to the 
squares of AC, CB, and twice the 
redangle BC,CD : To these equals 
add the square of BC, and the 

square 





OF EUCLID. 



57 



squares of AB, BC are equal to the square of AC, and twice ^^^kAL 
the sQuare of BC, and twice the redlangle BC, CD : But be- ^"^ 
cause BD is divided into two parts in C, the rectangle DB, 
BC is equaK to the re*Stangie.BC, CD and the square of BC : '3. 2- 
Arid the doubles of these are equal : Therefore the squares of 
AB, BC are equal to the square of AC, and twice the rect- 
angle DB, BC : Therefore the square of AC alone is less 
ihan the squares of AB, BC by twice the rectangle DB, BC. 

Lastly, let the side AC be perpendicular to a 

BC ; then is BC the straight line between the 
perpendicular and the acute angle at B ; and it 
is manifest, that the squares of AB, BC, are 
equals to the square of AC and twice the square 
of BC : Therefore, in every uiangle, &c. Q. 
E. D. 




«47. 1. 



PROP. XIV. PROB. 



X O describe a square that shall be equal to a given s<«^. 
rectilineal figure. 

Let A be the given redilineal figure ; it is required to de- 
scribe a square that shall be equal to A. 

Describe* the redtangular parallelogram BCDE equal to the ^^^ j 
redtilineal figure A. If then the sides of it BE, ED are equal to 
one another, it is a 
square, and what 
was required is 
now done : But if 
they are not equal, 
produce one of 
them BE to F, and 
make EF equal to 
ED and bisect BF 
in G : and from 

the centre G, at the distance GB, or GF, describe the semi- 
circle BHF, and produce DE to H, and join GH : Therefore 
because the straight line BF is divided into two equal parts ia 
the point G, and into two unequal at E, the re<^ngle BE, 
EF, together with the square gf EG, is equal^ to the square of ^' *• 
GF : But GF is equal to GH : therefore the redangle BE, EF, 




58 THE ELEMENTS, &c. 

BookII. together with the square of EG, is equal to the square of GH : 
T^^""^/ But the squares of HE, EG are equal*^ to the square of GH : 
''/■ ' Therefore the re6tangle BE, EF, together with the square of 
EG, is equal to the squares of HE, EG : Take away the 
square of EG, which is common to both ; and the remaining 
redlangle BE, EF is equal to the square of EH : But the 
redlangle contained by BE, EF is the parallelogram BD, be- 
cause EF is equal to ED ; therefore BD is equal to the square* 
of EH ; but BD is equal to the re<Slilineal figure A ; there- 
fore the redlilineal figure A is equal to the square of EH. 
Wherefore a square has been made equal to the given re<5lili- 
neal figure A, viz. the square described upon EH. Which 
was to be done. 



THL 



79 



ELEMENTS 



OF 



EUCLID. 



BOOK. III. 



DEFINITIONS. 

I. 

IliQUAL circles are those of which the diameters are "equal, 
<>r from the centres of which the straight lines to the circunv- 
ferences are equal. 

< This is not a definition, but a theorem, the truth of which 

* is evident ; for, if the circles be applied to one another, so 

* that their centres coincide, the circles must likewise Coin- 

* cide, since the straight lines from the centres are equal.' 

II. 
A straight line is said to touch 
a circle, when it meets the 
circle, and being produced 
does not cut it. 
III. 
Circles are said to touch one 
another, which meet but do 
not cut one another. 
IV. 
Straight lines are said to be equally dis- 
tant from the centre of a circle, when 
the perpendiculars drawn to them 
from the centre are equal. 
V. 
And the straight line on which the 
gr«ater perpendicular falls, is said to 
be farther from the centre. 




6(3 

Book III, 



THE ELEMENTS 
VI. 




A segment of a circle is the figure con- 
tained by a straight line and the cir- 
cumference it cuts ofF. 
VII. 

*' The angle of a segment is that which is contained by the 
" straight line and the circumference." 
VIII. 

An angle in a segment is the angle con- 
tained by two straight lines drawn 
from any point in the circumference 
of the segment, to the extremities of 
the straight line which is the base of 
the segment. 

IX. 

And an angle is said to insist or stand 
upon the circumference intercepted 
between the straight lines that contain the angle. 



X. 

The sector of a circle is the figure con-* 
tained by two straight lines drawn from 
the centre, and the circumference be- 
tween them. 




XI. 

Similar segments of a circle 
are those in which the an- 
gles are equal, or which 
contain equal angles. 




PROP. I. PROB. 

N- X O find the centre of a given circle. 

Let ABC be the given circle; it is, required to find itj 
centre. 

Draw within it any straight line AB, and bise£l* it in D ; 
from the point D draw '' DC at right angles to AB, and pro- 
duce it to E, and biseft CE.in F : The point F is the centre 
of the circle ABC. 

Foi 



» 10. 1. 



^ U. 1. 



OF EUCLID. 



6l 



For, if it be not, let, if possible, G be the centre, and join BooillL 
GA, GD, GB : Then, because DA is equal to DB, and DG ^•*^^'*^ 
common to the two triangles ADG, 
BDG, the two sides AD, DG are 
equal to the two BD, DG, each to 
each ; and the base GA is equal to the 
base GJ3, because they are drawn from 
the centre G* : Therefore the angle 
ADG is equal ' to the angle GDB : 
But when a straight line standing upon 
another straight line makes the adjacent 
angles equal to one another, each of the 
angles is a right angle"* : Therefore the 
angle GDB is a right angle: But FDB 

is likewibe a right angle; wherefore the angle FDB is equal 
to the anale GDB, the greater to the less, which is impos- 
sible : Therefore G is not the centre of the circle ABC . In 
the same manner it can be shewn, that no other point but F 
is the centre : that is, F is the centre of the circle ABC ; 
Which was to be found. 

;' Cor. From cnis it is manifest, that if in. a circle a straight 
line bisect another at right angles, the centre of the circle is in 
the line which biseds the other. 




<10Def;i. 



PROP. II. THEOR. 

1 F any two points be taken in the circumference 
of a circle, the straight line which joins them shall 
fall withhi the circle. 

Let ABC be a circle, and A, B any two points in thecircum> 
ference ; the straight line drawn from 
A to B shall fall within the circle. 

For, ifitdo not, let it fall, if possible, 
without, as AEB j find * D the centre 
of the circle ABCj and join AD, DB, 
and produce DF, any straight line 
meeting the circumference AB to E : 
Then because DA is equal to DB, the 
angle DAB is equal*" to the angle DBA ; 
and because AE, a side of the triangle 

■♦ N. B. Wbtaevcr the expressfba " straight lines from the centre," or " dra'wn 
from the e«nue," occurs, it is to b« understood that thej' are drava to the drcnm- 
fcrcnce, 

DAE, 




1.3. 



*«. 1. 



62 THEELEMENTS 

^ooK III. DAE, is produced to B, the angle DEB is greater "^ than the 
j'*?r^p^ angle DAE ; but DAE is equal to the angle DBE ; therefore 
the angle DEB is greater than the angle DBE : But to the 
*,19. 1, greater angle the greater side is opposite"^ ; DB is therefore 
greater than DE : But DB is equal to DF i wherefore DF is 
greater than DE, the less than the greater, which is impos- 
sible : Therefore the straight line drawn from A to B does 
not fall without the circle. In the same manner, it may be 
demonstrated that it does not fall upon the circumference ; it 
falls therefore within it. Wherefore, if any two points, &c, 
Q^E. D. 

PROP. III. THEOR. 

xF a straight line drawn through the centre of a 
circle hisect a straight line in it which does not pass 
through the centre, it shall cut it at right angles ; 
and if it cuts it at right angles, it shall bisect it. 

Let ABC be a circle; and let CD, a straight line drawn 
through the centre, bisedi any straight line AB, which does 
not pass through the centre, in the point F : It cuts it also at 
right angles. 

* '• ^' Take^ E the centre of the circle, and join EA, EB. Then, 

because AF is equal to FB, and FE common to the two tri- 
angles AFE, BFE, there are two sides in the one equal to two 
sides in the other, and the base EA is 
equal to the base EB j therefore the 

* 8. 1. angle AFE is equaP to the angle BFE: 

But when a straight line standing upon 

another niakes the adjacent angles tf qual 

to one another, each of them is a right 
'iODef.i. «^angle : Therefore each of the angles 

AFE, BFE is a right angle ; wherefore 

the straight line CD„ drawn through 

the centre bisecting another AB that 

does not pass through the centre, cuts 

the same at right angles. 

But let CD cut AB at right angles ; CD also bisects it, that 

is, AF is equal to FB. 

The same construction being made, because EA, EB from 
<5. 1. ^l'® centre are equal to one another, the angle EAF is equal** 

to the angle EBF : and the right angle AFE is equal to 

the right angle BFE : Therefore, in the two triangles, EAF, 
2 EBF^ 




OF EUCLID. 6,3 

EBF, there are two angles in one equal to two angles in the Booi in . 
other, and the side EF, which is opposite to one of the equal ^^"^'^^ 
angles in each, is common to both ; therefore the other sides • 26. 1. 
are equal'^i AF therefore is equal to FB. Wherefore, if a 
straight line, &c. Q^ E. D. 



PPOP. IV. THEOR. 

J.F in a circle two straight lines cut one another 
which do not both pass through the centre, they 
do not bisect each other. 

Let ABCD be a circle, and AC, BD two straight lines In 
it which cut one another in the point E, and do not both pass 
through the centre : AC, BD do not bisecft one another. 

For, if it is possible, let AE be equal to EC, and BE to ED: 
If one of the lines pass through the centre, it is plain that it 
cannot be bisected by the other which 
does not pass through the centre ; But 
if neither of them pass through the 
centre, take * F the centre of the cir- 
cle, and join EF : and because FE, a 
straight line through the centre, bi- 
sefls another AC which does not pass 
through the centre, it shall cut it at 
right'' angles J wherefore FEA is a 
right angle : Again, because the 
straight line FE bisedls the straight line BD which does not 
pass through the centre, it shall cut it at right'' angles ; where- 
fore FEB is a right angle : And FEA was shewn to be a right 
angle ; therefore FEA is equal to the angle FEB, the less to 
the greater, which is impossible : Therefore AC, BD do not 
biseS one another. Wherefore, if in a circle, &c. Q. E. D. 



PROP. V. THEOR. 

JlF two circles cut one another, they shall not have 
the same centre. 

Let the two circles ABC, CDG cut one another in the 
|)oints B, C i they have not tl^ same centre. 

For 




64 

Book III 



THE ELEMENTS 

For, if it be possible, let E be their centre ; join 
draw any straight line EFG meet- 
ing them in F and G ; and because 
E is the centre of the circle ABC, 
CE is equal to EF : Again, be- 
cause E is the centre of the circle 
CDG, CE is equal to EG : But 
CE was shewn to be equal to EF ; 
therefore EF is equal to EG, the 
less to the greater, which is im- 
possible : Therefore E is not the 
centre of the circles ABC, CDG. 
Wherefore, if two circles, &c. Q. E. D. 



EC, and 




PROP. VI. THEOR. 

JlF two circles touch one another internally, they 
shall not hav^e the same centre. 

Let the two circles ABC, CDE, touch one another inter- 
nally in the point C : They have not the same centre. 

For, if they can, let it be F; join FC and draw any straight 
lineFEB meeting them in E and B J 
And because F is the centre of the 
circle ABC, CF is equal to FB ; 
Also, because F is the centre of the 
circle CDE, CF is equal to FE : 
And CF was shewn equal to FB ; 
therefore FE is equal to FB, the less 
to the greater, which is impossible : 
V/herefore F is' not the centre of 
the circles ABC, CDE. Therefore, 
if two circles, &c. Q^ E. D. 




OF EUCLID. 65 

Book III. 

PROP. VII. THEOR. • — < — 

XF any point be taken in the diameter of a circle 
Avhich is not the centre, of all the straight lines 
which can be drawn froin it to the circumference, 
the greatest is that in which the centre is, and the 
other part of that diameter is the least; and, of any 
others, that which is nearer to the line which passes 
through the centre is always greater than one more 
remote : And from the same point there can be 
drawn only two straight lines that are equal to one 
another, one upon each side of the shortest line. 

Let ABCD be a circle, and AD its diameter, in which let 
any point F be taken which is not the centre : Let the centre 
be E ; of ail the straight lines FB, FC, FG, &c. that can be 
drawn from F to the circumference, FA is the greatest, and 
! FD, the other part of the diameter BD, is the least : And of 
the others, FB is greater than FC, and FC than FG. 

Join BE, CE, GE ; and because two sides of a triangle are 
greater^* than the" third, BE, EF are greater than BFj but AE *20. l. , 
is equal toEB ; therefore AE, tF, 
that is AF,is greater than BF: A- 
gain, because BE is equal to CE, 
and FE common to the triangles 
BEF, CEF, the twosidesBE,£F 
are equal to the two CE, EF i but 
the angle BEF is greater than the 

jangleCEF; therefore the base BF \ /^jiF\ / to^. 1. 

I is greater*" than the base FC : For 
! the same reason, CF is greater than 
' GF : Again, because GF, FE are 
' greater^ than EG, and EG is equal 

to ED ; GF, FE are greater than ED : Take away the com- 
mon part FE, and the remainder GF is greater than the re- 
mainder FD : Therefore FA is the greatest, and FD the least 
of all the straight lines from F to the circumference i and BF 
is greater than CF, and CF thaii GF. 

Also there can be drawn only two equal straight lines from 
the point F to the circumference, one upon each side of the 

F shortest 




66 THEELEMENTS 

Book III. shortest line FD : At the point E in the straight line EF, 
^^''^ make'' the angle FEH equal to the angle GEF, and join FH : 
Then because GE is equal to EH, and £F common to the 
two triangles GEF, HEF ; the two sides GE, EF are equal 
to thjC two HE, EF j and the angle GEF is equal to the an- 
'^ -i. 1. gle HEF ; therefore the base FG is equal"* to the base FH : 
5ut, besides FH, no other straight line can be drawn from F 
to the circumference equal to FG : For, if there can, let it be 
FK ; and because FK is equal to FG, and FG to FH, FK is 
equal to FH ; that is, a line nearertothat which passes through 
the centre, is equal to one which is more remote ; which is 
impossible. Therefore, if any point be taken, &c. Q. E. D. 



PROP. VIII. THEOR. 

If any point be taken without a circle, and straight 
lines be drawn from it to the circumference, where- 
of one passes through the centre; of those which fall 
upon the concave circumference, the greatest is that 
which passes through the centre ; and of the rest, 
that, which i§ nearer to that through the centre is al- 
ways greater than the more remote : But of those 
which fall upon the convex circumference, the least 
is that between the point without the circle and the 
diameter ; and of the rest, that which is nearer to 
the least is always less than the more remote : And 
only two equal straight lines can be drawn from the 
point into the circumference, one upon eacli side 
of the least. 

Let ABCbeacircle,and Danypoint without it, from which 
let the. straight lines DA, DE, DF,.DC be drawn to the cir- 
cumference, whereof DA passes through the centre. Of thos* 
which fall upon the concave part of the circumference AEFC, 
tiie greatest is AD which passes through the centre ; and the 
nearer to it is always greater than the more remote, viz. DE 
than DF, and DF than DC: But of those which fall upon the 
eonvcx circumference HLKG, the least is DG between the* 

I point 



I 



At. 



OF EUCLID. . 6f 

Book 111. 

Jjolnt D and the diameter AG ; and the nearer to it is always ^--v**-', 
less than the more remote, viz. DK than DL, and DL than 
DH. 

Take* M the centre of the circle ABC, and join ME,MF, ' i- 3. 
MC, MK, ML, MH : And because AM is equal to ME, add 
MD to each, therefore AD is equal to EM, MD i but EM,MD 
are greater^ than ED ; therefore also AD is greater than ED " 20. i. 
Again, because ME is equal to MF, and MD common to the 
triangles EMD, FMD ; EM, MD 
are equal to FM, MD ; but the 
angle EMD is greater than the 
angle FMDj therefore the base 
ED is greater'^ than the base FD : 
In like manner it may be shewn 
that FD is greater than CD : 
Therefore DA is the greatest ; and 
DE greater than DF, and DF than 
DC : And because MK, KD are 
greater'' than MD, and MK is 
equal to MG, the remainder KD 
' is greater ^ than the remainder 
! GD, that is GD is less than KD : 
■ And because MK, DK are drawn 
to the point K within the triangle 
MLD from M, D, the extremi- 
ties of its side MD, MK, KD are 
Jess = than ML, LD, whereof MK 

is equal to ML j therefore the remainder DK is less than the 
remainder DL : In like manner it may be shewn, that DL is 
less than DH : Therefore DG is the least, and DK less than 
DL, and DL than DH : Also there can be drawn only two 
equal straight lines from the point D to the circumference, 
j one upon each side of the least : At the point M, in the straight 
line MD, make the angle DMB equal to the angle DMK, 
land join DB : And because MK is equal to MB, and MD 
I common to the triangles KMD, BMD, the two sides KM, 
|MD are equal to the two BM, MD ; and the angle KMD 
lis equal to the angle BMD ; therefore the base DK is equaK ^ *• ^• 
to the base DB : But, besides DB, there can be no straight 
jline drawn from D to the circumference equal to DK : For, 
jif there can, let it be DN j and because DK is equal to DN, 
iand also to DB ; therefore DB is equal to DN, that is, the 
,nearer to the least equal to the more remote, which is im- 
ipossible. If therefore, any point, &c. Q^ E. D, 
1 F2 ■ ' 




21.1. 



68 

Book III. 



THE ELEMENTS 



PROP. IX. THEOR. 



»7. 



J[ F a point be taken within a circle, from which there 
fall more than two equal straight lines to the cir- 
cumference, that point is the centre of the circle. 

Let the point D be taken within the circle ABC, from 
which to the circumference there fall more than two equal 
straight lines, viz. DA, DB, DC, the point D is the centre of 
the circle. 

For, if not, let E be the centre, 
join DE and produce it- to the cir- 
cumference in F, G : then FG is 
a diameter of the circle ABC : And 
because in FG, the diameter of the 
circle ABC, there is taken the 
point D, which is not the centre, 
DG shall be the greatest line from it 
to the circumference, and DC great- 
er » than DB, and DB than DA : 
But thej^ are likewise equal, which 
is impossible ; Therefore E is not 

the centre of the circle ABC: In like manner, it may be demon- 
strated, that no other point but D is the centre ; D therefore 
is the centre. Wherefore, if a point be taken, &c. Q. E. D, 




PROP. X. THEOR. 

i^NE circumference of a circle cannot cut anothe 
in more than two, points. 

If it be possible, let the circumfe- 
rence FAB cut the circumference 
DEF in more than two points, viz. 
in. B, G, F ; take the centre K of the ^ 
cirrk- ABC, and join KB, KG, KF: 1^ 
And because within the circle DEF 
there is taken the point K, from 
which to the circumference DEF 
fall more tlian two equal straight 
lines KB, KG, KF, the point K is'" 




OF EUCLID."- 6^ 

the centre of the circle DEF : But K is also the centre of the Boo* iii. 
circle ABC; therefore the same point is the centre of two cir- 
cles that cut one another, which is impossible^ Therefore " 5. C\ 
one circumference of a circle cannot cut another in more than 
^wo points. Q_ E. D. 

PROP. XI. THEOR. 

X F two circles touch each other internally, the 
straight line which joins their centres heing pro- 
iduced shall pass through the point of contact. 

Let the two circles ABC, ADE touch each other internally 
in the point A, and let F be the centre of the circle ABC, ar.d 
G the centre of the circle ADE : The 
straight line which joins the centres 
F, G, being producedj passes through 
the point A. 

For, if not, let it fall otherwise, if 
possible, as FGDH, and join AF, 
AG: And because AG, GF are great- 

ec * than FA, that is, than FH, for \ \^ ^ ^^^^^ ' *°' ^' 
FA is equal to FH, both being from 
the same centre ; take away the com- 
mon part FG; therefore the remain- 
der AG is greater than the remainder GH; But AG is equal 
to GD ; therefore GD is greater than GH, the less than the 
greater, which is impossible. Therefore the straight line 
whichjoins the points F, G cannot fall otherwise than upon the 
point A, that is, it must pass through it. Therefore, if two 
circles, Sec. Q^ E. D. 

PPOP. XII. THEOR. 

J.F two circles touch each other externally, the 
straight line which joins their centres shall pass 
through the point of contact. 

Let the two circles ABC, ADE, touch each other exter- 
nally in the point A ; and let F be the centre of the circle 
ABC, and G the centre of ADE : The straight line which 
joins the points F, G shall pass through the point of contact A. 

For, if not, let it pass otherwise, if possible, as FCDG, and 
F 3 join 




7« 



THE ELEMENTS 




B«ojc in. join FA, AG: And because F is the centre of the circls ABa 
^^''"^ AF is equal to FC : Also 
because G is thecentreof 
' the circle ADE, AG is 
equal to GD: Therefore 
FA, AG are equal to FCj 
DG i wherefore the 
whole FG is greater than 
FA, AG : But it is also 

•20. I, less*; which is impossi- 
ble : Therefore the straight line which joins the points F, G 
shall not pass otherwise than through the point of contact A, 
that is, it must pass through it. Therefore, if two circles, 
&c. Q. E. D. 

PROP. XIII. THEOR. 

vJNE circle cannot touch another in more points 
than one, whether it touches it on the inside or 
outside. 

For, if it be possible, let the circle EBF towch the circle 

ABC in more points than one, and first on the inside, in the 

« 10.11, 1. points B, D ; join BD, and draw* GH bisecting BD at right 

angle.s ^.Therefore because the points B, D are in the circum- 



H B 





"1.3. 
•^Cor. 1.3, 
" 11, 3. 



ference of each of the circles, the straight line BD falls within 
each'' of them : And their centres are<^ in the straight line GH 
which bisedts BD at right angles : Therefore GH passes 
through the point oFcontacf*; but it does not pass through it, 
because the points B, D are without the straight line GH, 
which is absurd : Therefore one circle cannot touch another 
on the inside in more points than one. 

Nor can two circles* touch one another on the outside in 

more 



OF EUCLID. 



7» 



more than one point ; For, if it be possible, letthecircle ACK Book jH. 
touch the circle ABC in the points A, C, and join AC : There- ^'^^''^^ 
fore, because the two points A, C are in 
the circumference of the circle ACK, 
the straight line AC which joins them 
shall fall within** the circle ACK : And 
the circle ACK is without the circle 
ABC J and therefore the straight line 
AC is without this last circle ; but be- 
cause the points A, C are in the circum- 
ference of the circle ABC, the straight 
line AC must be within'' the same cir- 
cle, which is absurd : Therefore one 
circle cannot touch another on the out- 
side in more than one point : And it 
has been shewn, that they cannot touch on the inside in more 
points than one. Therefore, one circle, &c. Q. E. D. 




PROP. XIV. THEOR. 

JCuQUAL straight lines in a circle are equally dis- 
tant from the centre; and those which are equally 
distant from the centre, are equal to one another. 

Let the straight lines AB, CD, in the circle ABDC, be e- 
qual to one another j they are equally distant from the centre. 

TakeE the centre of the circle ABDC, and from it draw EF, 
EG perpendiculars to AB, CD : Then, because the straight line 
EF, passing through the centre, cuts thsstraight line AB, which 
does not pass through the centre, at 
right angles, it also bisefls' it : Where- 
fore AF is equal toFB, and AB double 
of AF. For the same reason CD is 
double of CG : And AB is equal to 
CD ; therefore AF is equal to CG ; 
And because AE is equal to EC, the 
square of AE is equal to the square of 
EC -. But the squares of AF, FE are 
equal'' to the square of AE, because 
the angle AFE is a right angle; and 
for the like reason, the squares of EG, GC are equal to the 
square of EC : Therefore the squares of AF, FE are equal to 
the squares" of CG, GE, of which the square of AF is equal to 

F 4 the 




72 



THE ELEMENTS 



^°^*I"- the square of CG i because AF is equal to CG ; therefore the 
remaining square of FE is equal to the remaining square of 
EG, and the straight line EF is therefore equal to EG : But 
straight lines in a circle are said to be equally distant from the 
centre, when the perpendiculars drawn to them from the cen- 

•4.def, 3. tre areequal'^: Therefore AB, CD be equally distant from 
the centre. 

Next, if the straight lines AB, CD be equally distant from 
the centre, that is if FE be equal to EG; AB is equal to CD: 
For, the same construction being made, it may, as before, be 
demonstrated, that AB is double of AF, and CD double of 
CG, and that the squares of EF, FA are equal to the squares 
of EG, GC ; of which the square of FE is equal to thesquarc 
of EG, because FE is equal to EG ; therefore the remaining 
square of AF is equal to the remaining square of CG; and the 
Straight line AF is therefore equal to CG: And AB is double 
of AF, and CD double of CG ; wherefore AB is equal to 
CD. Therefore equal straight lines, &c. Q^ E. D. 



PROP. XV. THEOR. 



SeeN. 



10. I. 



1 HE diameter is the greatest straight line in a cir- 
cle ; and, of all others, that which is nearer to the 
centre is always greater than one more remote : and 
the greater is nearer to the centre than the less. 

Let ABCD be a circle, of which 
the diameter fs AD, and the centre E ; 
and let BC be nearer to the centre 
than FG ; AD is greater than any 
straight line BC which is not a dia- 
meter, and BC greater than FG. 

Fi-om the centre draw EH, EK per- 
pendiculars to BC, FG, and join EB, 
EC, EF ; and because AE is equal to 
EB, and ED to EC, AD is equal to 
EB, EC ; but EB, EC are greater •' 
than BC : wherefore, also AD is 
greater than BC. 

Apd, because BC is nearer to the cehtre than FG, EH Is 

less 




OF EUCLID. 73 

less^ than EK . But, as was demonstrated in the preceding. Book IH. 
BC is double of BH, and FG double of FK, and the squares of e ^ ^^ 3^ 
EH, HB are equal to the squares of EK, KF, of which the 
square of EH is less than the square of EK, because EH is 
less than EK ; therefore the square of BH is greater than the 
square of FK, and the straight line BH greater than FK. and 
therefore BC is greater than FG. 

Next, Let BC be greater than FG j BC is nearer to the cen- 
tre than FG, that is, the same construction being made, EH is 
less than EK : Because BC is greaterthan FG, BH likewise is 
greater than KF : And the squares of BH, HE are equal to the 
squares of FK, KE, of which the square of BH is greater than 
the square of FK, because BH is greater than FK ; therefore 
the square of EH is less than thesquare of EK, and the straight \ 
lilieEHless than EK. Wherefore the diameter, &c, Q^E. D. 

PROP. XVI. THEOR. 

JL HE straight line drawn at right angles to thedia- seeN. 
meter of a circle, from the extremity of it, falls 
without the circle; and no straight linecanhe drawn 
between that straightlineand the circumference from 
the extremity, so as not to cut the circle ; or, which 
is the same thing, no straight line can make so great 
an acute angle with the diameter at its extremity, 
or so small an angle with the straight line, which is 
at right angles to it, as not to cut the circle. 

Let ABC be a circle; the centre of which is D, and the dia- 
meter AB : the straight line drawn at right angles to AB from 
its extremity A, shall fallwithout the circle. 

For, if it does not, let it fall, if 
possible, within the circle, as AC, 
and draw DC to the point C where 
it meets the circumference : And 
because DA is equal to DC, the 
angle DAC is equal^ to the angle 
ACD; but DAC is a right angle, 
therefore ACD is a right angle, 
and the angles DAC, ACD are 

therefore equal to two right angles j which is impossible ^ : 

Therefore 




u 



THE ELEMENTS 



' 12. 1. 

* i?. 1, 



Book iu. Therefore the straight line drawn from A at right angles to. 

^•^'^^'^^ j8A does not fall within the circle: In the same manner, it 
may be demonstrated, that it does not fall upon the circum- 
ference; therefore it must fall without the circle, as AE. 

And between the straight line AE and the circumference no 
straight line can be drawn from the point A which -does not 
cut the circle : For, if possible, let FA be between them, and 
from the point D draw '^ DG perpendicular to FA, and let it 
meet the circumference in H : And because AGO is a right 
angle, and DAG less ^ than aright angle, DA is greater'^ than 
DG: But DA is equal to DH : 
therefore DH is greater than DG, 
the less than the greater, which is 
iitipossible : Therefor'^ no straight 
line can be drawn from the point 
A between AE and the circumfe- 
rence, which does not cut the cir- 
cle, or, which amounts to the same 
thing, however great an acute angle 
a straight line makes with the dia- 
meter at the point A, or however 
small an angle it makes with AE, 
the circumference passes between that straight line and thff 
perpendicular AE. * And this is all that is to be understood, 

* when, in the Greek text, and translations from it, the angld 

* of the semicircle is said to be greater than any atute reitili-.. 

* ncal angle, and the remaining angle less than any rectilineal 

* angle.' Q. E. D. 

Cor. From this it is manifest, that the straight line which 
is drawn at right angles to the diameter of a circle from the ex-, 
tremity of it, touches the circle ; and that it touches it only in 
one point, because, if it did meet the cifcle in two, it would 
• 2. 3. fall within it' . * Also it is evident that there can be but one 
' straight line which teuches the circle in the same point. * 




PROP. XVII. PROB. 

l O draw a stiaight line from a given point, either 
.witkout or iu the circumference, which sl^ll touch 
a given circle. 
First, let A be a given point without the given circle BCDj 



OF EUCLID. 



75 



» 1.3, 



xil. 1. 



it is required to draw a straight line from A which shall touch ^^^^^' 
the circle. 

Find * the centre E of the circle, and join AE ; and from 
the centre E, at the distance EA, describe the circle AFGj 
from the point D draw '' DF at right angles to EA, and join 
EBF, AB. AB touches the circle BCD. 

Because E is the centre 
of the circles BCD, AFG, 
EA is equal to EF : And 
ED to EB ; therefore the 
two sides AE, EB are equal 
to the two FE, ED, and 
they contain the angle at 
E common to the two tri- 
angles AEB, FED ; there- 
fore the base DF is equal 
to the base AB ; and the 
triangle FED to the trian- 
gle AEB, and the other angles to the other angles'^ : There- «4. i. 
fore the angle EBA is equal to the angle EDF : But EDF is 
a rightanglcjWhereforeEBA.isaright angle: And EB is drawn 
from the centre : But a straight line drawn from the extremi- 
ty of a diameter, at right angles to it, touches the circle<* : *""' ' 
Therefore AB touches the circle ; and it is drawn from the 
given point A. Which was to be done. 

But if the given point be in the circumference of the circle, 
as the point D, draw DE to the centre E, and DF at right 
angles to DE ; DF touches the circle"^. 




PROP. XVIII. THEOR. 



JLF a straight line touches a chcle, the straight line 
drawn from the centre to the point of contact, shall 
be pei-pendicular to the line touching the circle. 

Let the straight line DE touch the circle ABC in the point 
Ci take the centre F, and draw the straight line FC : FC is 
perpendicular to DE. 

For, if it be not, from the point F draw FBG perpendicular 
to D£ J and because FGC is a right angle, GCF is** an acute b 17 j 
angle i and to the greater angle the greatest"^ side is opposite : « 19! \. 

Therefore 



76 



THE ELEMENTS 



^^^2^^2!L' Therefore FC is greater than FG ; 

^^"^^^^^ but FC is equal to FB ; therefore 
FB is greater than FG, the less 
than the greater, which is impos- 
sible : Wherefore FG is not per- 
pendicular to DE : In the same 
manner it may be shewn, that no 
other is perpendicular to . it besides 
FC, that is FC is perpendicular to 
DE» Therefore, if a straight line, 
&c. Q- E. D. 




G E 



PROP. XIX. THEOR. 

If a straight line touches a circle, and from thr 
point of contact a straight line be drawn at right 
angles to the touching Une, the centre of the circle 
shall be in that line. 

Let the straight line DE touch the circle ABC In C, and 
from C let CA be drawn at right angles to DE j the centre 
»f the circle is in CA. 

For, if not, let F be the centre, if possible, and join CF j 
Because DE touches thecircle ABC, 
and FC is draw-n from the centre to 
the point of contact, FC is perpendi- 
' IS. 3. cular * to DE ; therefore FCE is a 
right angle : But ACE is also a right 
angle ; therefore the angle FCE is 
equal to the angle ACE, the less to t^\^ 
the greater, which is impossible : 
Wherefore F is not the centre of the 
circle ABC: In the same manner, it 
may be shewn, that no 'other point 
which it not in CA, is the centre ; that is, the centre is in C A , 
Therefore, if a straight line, &c. Q. E. D. 

prop'. XX. THEOR. 

',2. N. 1 HE angle at the centre of a circle rs double of the 
angle at the circumference, upon the same base, 




that is, upon the same part of the circumference 



Let 



OF EUCLID. 



77 




»5. 1. 



*3^1. 



Let ABC be a circle, and BEC an angle at the centre, and Book ill. 
BAC an angle at the circuniference, which have the same cir- ^■'•**''*^ 
cumference BC for their base •, the angle 
BEC IS double of the angle BAC. 

First, let E, the centre of the circle, 
be within the angle BAC, and join AE, 
and produce it to F : Because EA is equal 
to EB, the angle EAB is equal'' to the 
angle EEA; therefore the angles EAB, 
EBA are double of the angle EAB j but 
the angle BEF is equal'' to the angles 
EAB, EBA ; therefore also the angle 
BEF is double of the angle EAB : For 
the same reason, the angle FEC is double of the angle E AC : 
Therefore the whole angle BEC is doubleof the whole angle 
BAC. 

Again, let E, the centre of the 
circle, be without the angle BDC, and 
join DE, and produce it to G. It 
may be demonstrated, as in the first 
case, that the angle GEC is double 
of the angle GDC, and that GEB, a 
part of the first, is double of GDB, a 
part »f the other ; therefore the re- Gr 
maining angle BEC is double of the 
remaining angle BDC. Therefore the 
angle at the centre, &c. Q^E. D. 




PROP. XXI. THEOR. 

JL HE angles in the same segment of a circle are 
equal to one another. 

Let ABCD be a circle, and BAD, 
BED angles in the same segment 
BAED : l^he angles BAD, BED are 
equal to one another. 

Take F, the centre of the circle 
ABCD: And, first, let the segment. 
BAED be greater than a semicircle, 
and join BF, FD : And because the 
angle BFD is at the centre, and the 
angle BAD at the circumference, 
a;)d that they have the same part of 

the 



SesK. 




Book III, 
» 20. 3. 



THE ELEMENTS 

the circumference, viz. BCD for their base ; therefore the an- 
gle BFD is double* of the angle BAD : For the same reason, 
the angle BFD is double of the angle BED: Therefore the 
angle BAD is equal to the angle BED. 

But, if the segment BAED be not greater than a semicircle, 
let BAD, BED be angles in it ; these . -j, 

also are equal to one another : Draw Jk-^j^ 

AF to the centre, and produce it to 
C, and join CE : Therefore the seg- "A 
ment BADC is greater than a semi- 
circle ; and the angles in it BAG, 
BEC are equal, by the first case : For 
the same reason, because CBED is 
greater than a semicircle, the angles 
CAD, CED are equal : Therefore 
the whole angle BAD is equal to the 
whole angle BED. Wherefore the angles in the same seg- 
ment, &c. Q. E. D. 




PROP. XXII. THEOR. 



•32. 1. 



*21.: 



X HE opposite angles of any quadrilateral figure 
described in a circle, are together equal to two 
right angles. 

Let ABCD be a quadrilateral figure in the circle ABCD ; 
any two of its opposite angles are together equal to two right 
angles. 

Join AC, BD ; and because the three angles of every tri- 
angle are equal* to two right apgles, the three angles of the 
triangle CAB, viz. the angles CAB, ABC, BCA are equal to 
two right angles : But the angle CAB j) 

is equal^ to the angle CDB, because v'-'''7"^^"<:>v r 
they are in the same segment BADC, /^ / \ ^'^'-' 
and the angle ACB is equal to the 
angle ADB, because they are in the 
same segment ADCB : Therefore the 

whole angle ADC is equal to the an- "\ /B 

gle CAB, ACB: To each of these 
equals add the angle ABC ; therefore 
the angles ABC, CAB, BCA are 
equal to the angles ABC, ADC : But ABC, CAB, BCA are 
equal to two right angles ; therefore alsothe angles ARC, ADC 
are equal to two right angles : In the same manner, the angles 

BAD 




OF EUCLID. 79 

BAD, DCB, may be shewn to be equal to two right angles. Boo» in. 
Therefore, the opposite angles, &c. Q^ E. D. >*'n'^«' 



PROP. XXIII. THEOR. 

U PON the same straight line, and upon the same see k. 
side of it, there cannot be two similar segments of 
circles, not coinciding with one anotiier. 

If it be possible, let the twe similar segments of circles, viz. 
ACB, ADB, be upon the same side of the same straight line 
AB, not coinciding with one another: Then, because the cir- 
cle ACB cuts the circle ADB in the 
two points A, B, they cannot cut one 
another in any other point* : One of 
the segments must therefore fall within 
the other : Lee ACB fall within ADB, 
and draw the straight line BCD, and 
join CA, DA : And because the seg- 
ment ACB is similar to the segment ADB, and that similar 
segments of circles contain'' equal angles; the angles ACB is ^iLdef-S. 
equal to the angle ADB, the exterior to the interior, which is 
impossible.*^ rherefore, there cannot be two similar seg^ <= J6. i. 
ments of a circle upon the same side of the same line, which 
do not coincide. Q^ E. D. 




• 10. 3. 



PROP. XXIV. THEOR. 

OlMILAR segments of circles upon equal straight Sm n. 
lines, are equal to one another. 

Let AEB, CFD be similar segments of cricles.upon-the 
equal straight lines AB, CD j the segment AEB is equal t# 
the segment CFD, 

For if the seg- 
ment AEB be 
applied to the 
segments CFD, 

so as the point A 

be on C, and A B C ~ D 

the straight line 

AB upon CD, the point B shall x:oincide with the point 9, be- 
cause 





8o 



THE ELEMENTS 



Book in, cauSc AB is equal te CD ; Therefore the straight line AB co- 
•^23?3/ inciding with CD, the segment AEB must ^ coincide with 

the segment CFD, and therefore is equal to it. Wherefore 

similar segments, &cc. Q. E. D. 



SeeN. 



«10. 1. 



6. 1. 



"9.3. 



PROP. XXV. PROB. 
A SEGMENT of a circle beinsr oiven, to describe 



'» &■ 



the circle of M^hich it is the segment. 

Let ABC be the given segment of a circle ; it is required 
to describe the circle of which it is the segment. 

Bised* AC in D, and from the point D draw** DB at right 
angles to AC, and join AB : First, let the angles ABD, BAD 
be equal to one another ; then the straight line BD is equal*^ 
to DA, and therefore to DC ; and because the three straight 
lines DA, DB, DC, are all equal j D is the centre of the cir-^ 
cle.'' From the centre D, at the distance of any of the three 
DA,DB,DC,describe a circle ; this shall pass through the other 
points i and the circle of which ABC is a segment is described j 
And because the centre D is in AC, the segment ABC is a se- 




23. 1. 



■*. 1. 




micircle : But if the angles ABD, BAD are not equal to one 
another,at the point A, in the straight lineAB make'= the angle 
BAE equal to the angle ABD, and produce BD,if necessary, to 
E,andjoinEC: And because the angle ABE is equal to the angle 
BAE, the straight line BE is equal '^ to EA : And because AD 
is equal to DC, and DE common to the triangles ADE, CDE, 
the two sides AD, DE are equal to the two CD, DE, each to 
each J and the angle ADE is equal to the angle CDE, for 
each of them is a right angle; therefore the base AE is equal 
'^to the base EC : But AE was shewn to be equal to EB, where- 
fore also BE is equal to EC : And the three straight lines AE, 

EB, 



OF EUCLtD. Si 

EB, EC are therefore equal to one another ; wherefore*^ E is ^^^- 
the centre of the circle. From the centre E, at the distance of 
any of the three AE, EB, EC, describe a circle, this shall pass " 9- 3. 
through the other points j and the circle of which ABC is a seg- 
ment is described: And it is evident, that if the angle ABD be 
greater than the angle BAD, the centre E falls without the 
segment ABC, which therefore is less than a semicircle : But 
if the angle ABD be less than BAD, the centre E falls withia 
the segment ABC, which is therefore greater than a semicircle: 
Wherefore a segment of a circle being given, the circle is de- 
scribed of which it is a segment. Which was to be done. 



PROP. XXVI. THEOR. 

J. N equal circles, equal angles stand upon equal 
circumferences, whether they be at the centres or 
circumferences. 

Let ABC, DEF be equal circles, and the equal angles BGC, 
EHF at their Centres, and BAC, EDF at their circumferences: 
The circumference BKC is equal to the circumierence ELF. 

Join BC, EF j and because the circles AB J, DEF are equal, 
the straight lines drawn from their centres are equal : There- 
fore the two sides BG, GC, are equal to the two EH, HF ; 




md the angle at G is equal to the angle at H ; therefore the 
l)ase BC is equal* to the base EF. And because the angle at A » 4. i. 
s equal to the angle at D, thesegment BAC is similar'' to the ''-^^•«*<.7' 
Kgment EDF ; and they are upon equal straight lines BC, EF j 
3ut similar segments of circles upon eqiaai straight lines are 
:qual<= to one another, therefore the segment BAC is equal to c ^i. 3. 
ihe segment EDF: But th? whole circle ABC is equ^l to the 

G whole 



82 



THE ELEMENTS 



Book III. whole DEF , therefore the remaining segment BKC is equal 
'^^'^^'''^ to the remaining segment ELF, and the circumference BKC 

to the circumference ELF. Wherefore, in equal circles, Sec. 

Q. E. D. 



PROP. XXVn. THEOR. 



»2. 8. 



XN equal circles, the angles which stand upon 
equal circumferences are equal to one another, 
AV'hether they be at the centres or circumferences. 

Let the angles BGC, EHF at the centres, and BAG, EDF at 
the circumferences of the equal circles ABC, DEF stand upon 
the equal circumferences BC, EF : The angle BGC is equal 
to the angle EHF, and the angle BAC to the angle EDF. 

If the angle BGC be equal to the angle EHF, it is manifest 
* that the angle BAC is also equaj to EDF. But, if not, one 



23. 1. 



* 2». 3. 




of- them is the greater : Let BGC be the greater, and at the 
point G, in the straight line BG, make*" the angle BGK equal 
to the angle EHF; but equal angles stand upon equal circum- 
ferences% when they are at the centre; therefore the circura-, 
ference BK is equal to the circumference EF : But EF is equaJ 
to BC; therefore also BK is equal toBC, the less to the greater, 
which is impossible: Therefore the angle BGCis not unequal 
to the angle EHF ; that is, it is equal to it : And the angle at 
A is half of the angle BGC, and the angle at D half of th( 
angle EHF : Therefore the angle at A is equal to the angl< 
at D. Wherefore, in equal circles. &c, C^ E. P. 



> 



OF EUCLID. 83 

Boon III. 

PROP. XXVIII. THEOR. 

In equal circles, equal straight lines cut off equal 
circumteiences, the greater equal to the greater, 
and the le88 to the less. 

Let ABC, DEF be equal circles, and BC, EF equal straight 
lines in them, which cut ofF the two greater circumferences 
BAG, EDF, and the two less BGC, EHF : the greater BAC is 
equal to the greater EDF, and the less BGC to the less EHF. 

Take^K,L, the centresofthecircles, and join BK,KC,EL, » 1. 2. 
LF : And because the circles are equal, the straight lines from 




Cr 

their centres are equal ; therefore BK, KC are equal to EL, 
LF ; and the base BC is equal to the base EF ; therefore the 
angle BKC is equal'' to the angle ELF : But equal angles stand « s. l. 
upon equal"^ circumferences, when they are at the centres j c ^g. ^ 
therefore the circumference BGC is equal to the circumfe- 
rence EHF. But the whole circle ABC is equal to the whole 
EDF J the remaining part therefore of the circumference, viz. 
BAC, is equal to the remaining part BDF. Therefore, ia 
equal circles, &c. Q^ E. D. 

PROP. XXIX. THEOR. 

N equal circles, equal circumferences are sub- 
tended by equal straight lines. 

Let ABC, DEF be equal circles, and let the circumferences 
BGC, EHF also be equal j and join BC, EF ; The straight 
line BC is equal to the straight line EF. 

G 2 Take 



84 THE ELEMENTS 

Book III. Tak€^ K, L, the centres of the circles, and join BK, KC, 
^J'g'^^ EL, LF : And because the circumference BGC is equal to the 



27. 



'^i. 1. 




circumference EHF, the angle BKC is equal*" to the angle 
ELF : And because the circles ABC, DEF, are equal, the 
straight lines from their centres are equal : Therefore BK, 
KC are equal to EL, LF, and they contain equal angles : 
Therefore the base BC, is equal"^ to the base EF. There- 
fore, in equal circles, &c. Q. E. D. 



PROP. XXX. PROB. 

X O bisect a given circumference, that is, to du 
vide it into 4wo equal parts. 

Let ADB be the given circumference ; it is required to 
sect it. 

* 10. 1. Join AB, and bisect^ it in C ; from the point C draw 

at right angles to AB, and join AD, DB : The circumfa- 
rence ADB is bisected in the point D. 

Because AC is equal toCB, and CD common Ui the triangles 
ACD, BCD, the two sides AC, CD 
are equal to the two BC, CD ; and 
the angle ACD is equal to the angle 
BCD, because each of them is a right 
angle : Therefore the base AD is equal 

"4. 1. ^ to the base BD. But equal straight 

« 28. 3. lines cut off equal<^ circumferences, the greater equal to the 
greater, and the less to the less, and AD, DB are each of them 
less than a semicircle ; because DC passes through the cen- 

< Ccr. 1. 3. ^^^^ ' Wherefore the circumference AD is equal to tl?e cir- 
cumference DB : Therefore the given circumference is bi- 
sected in D, Which was to be done. 




11 



OF EUCLID. 85 

Book III. 

PROP. XXXI. THEOR. '^^'^ 

1 X a circle, the angle in a semicircle is a right an- 
gle ; but the angle in a segment greater than a se- 
micircle is less than a right angle ; and the angle in ' 
a segment less than a semicircle is greater than a 
right angle. 

Let ABCD be a circle, of which the diameter is BC, and 
centre E ; and draw CA, dividing the circle into the segments 
ABC, ADC, and join BA, AD, DC ; the angle in the semi- 
circle BAC is a right angle; and the angle in the segment 
ABC; which is greater than a semicircle, is less than a right 
angle ; and the angle in the segment ADC, which is less than 
a semicircle, is greater than a right angle. 

Join AE, and produce BA to F : and because BE is equal 
to EA, the arigle EAB is equal^ to EBA ; also, because AE '"'■ '" 
is equal to EC ; the angle EAC is 
equal to ECA ; wherefore ths 
whole angle BAC is equal to the 
two angles ABC, ACB: But FAC, 
the exterior angle of the triangle 
ABC, is equal'' to the two angles 
ABC, ACB : therefore the angle 
BAC is equal to the angle FAC, 
and each of them is therefore a 
righf^ angle : Wherefore the angle 

BAC in a semicircle is a right an- ^^ >/ ^lo. dcf. 1. 

g!e. 

And because the two angles ABC? BAC of the triangle 
ABC are together less"^ than two right angles, and that BAC 
s a right angle, ABC must be less than a right angle ; and 
:herefore the angle in a segment ABC greater than a semicir- 
:le, is less than a right angle. 

And because ABCD is a quadrilateral figure in a circle, any 
wo of its opposite angles are equal'' to two right angles :- < 
:herefore ths angles ABC, ADC are equal to two right an-* 
^es ; and ABC is less than a right angle ; wherefore the other 
kDC is greater than a right angle. 

Besides, it is manifest, that the circumference of the greater 
segment ABC falls without the right angle CAB ; but the 
arcumference of tb€ less segment ADC faUs within the right 

gle CAP.' ' And this is all that is meant, when in the 
G 3 « Greek 




36 

Book III. 



THE ELEMENTS 

' Greek text; and the translations from it, the angle of the 
' greater segment is said to be greater, and the angle of the less 
' segment is said to be less, than a right angle.' 

Cor. From this it is manifest, that if one angle of a triangle 
be equal to the other tv;o, it is a right angle, because the angle 
adjacent to it is equal to the same two; and when the adjacent 
angles ace equal, they are right angles. 



PROP. XXXII. THEOR. 



» :;. \. 



" 1?. 3. 

■=31. 3. 
*32. 1. 



'i::. 3. 



F a straight line touches a circle, and from the 
point of contact a straight line he drawn cutting 
the circle, the angles made by this line with the line 
touching the circle, shall be equal to the ^angles 
which are in the alternate segments of the circle. 

Let the straight line EF touch the circle ABCD in B, and 
from the point B let the straight Ijne BD be drawn, cutting 
the circle: The angles which BD makes with the touching 
line EF shall be equal to the angles in the alternate segmenis 
of the circle : that is, the angle FBD is equal to the angle 
which is in the. segment DAB, and the angle DBE to the aa- 
G^le in the se,^ment BCD. 

From the point B draw^' BA at right angles to EF, and taiaj 
any point C in the circumference BD, and join AD, DC, CB 
and because the straight line EF touchts the circle ABCDi; 
the point B, and BA is drawn at 
right angles to the touching line 
from the point of contact B, the 
centre of the circle is^ in BA ; 
therefore the angle ADB in a semi- 
circle is a righf^ angle and con<^e- 
quently theother two angles BAD, 
ABD are equal'^ to a right angle : 
But ABF is likewise a right angle : 
therefore the angle ABF is equal to 
the angles BAD, ABD: Pake from 
these equals the common angle 

ABD ; therefore the remaining dngle DBF is equal to the an- 
gle BAD, which is in the alternate segment of tlje circle j and 
because ABCD is a quadrilateral figure in a circle, the opposit^ 
angles BAD, BCD are equal*" to two right angles : therefo 

ths 




OF EUCLID. 



«7 



the angles DBF, DBE, being likewise equaF to two right an- ^^^ in. 
gles, are equal to the angles BAD, BCD : and DBF has been f**^7*^ 
proved equal to BAD : Therefore the remaining angle DBE 
is equal to the angle BCD in the alternate segment of the cir- 
cle. Wherefore, if a straight line, &c. Q. £. D. 



PROP. XXXIII. PROB. 

U POX a ariven straio-btline to describe a seo-ment 
ot a circle, containing an angle equal to a given ^'=*^^'- 
rectilineal angle. 



Let AB be the given straight line, and the angle at C the 
given rectilineal angle ; it is required to describe upon the 
given straight line AB a segment of a circle, containing an 
angle equal to the angle C. 

First, let the angle at C be a right 
angle, and bisect* AB in F, aad 
from the centre F, at the distance 
FB, describe the semicircle AHB ; 
therefore the angle AHB in a semi- 




10. 1. 



«" 31. 3. 



Ml.l. 



circle is*" equal to the right angle 
at C. 

But, if the angle C be not a right angle, at the point A, in 
the straight line AB, make"^ the angle BAD equal to the angle '23. i. 
C, and from the point A 

draw'^ AE at right angles to ^^ 

ADi bisect^ AB in F, and 
from F draw^ FG at right 
angles to AB, and join GB : 
And because AF is equal to 
FB, and FG common to the 
triangles AFG, BFG, the 
two sides AF, FG are equal 
to the two BF, FG ; and the 
angle AFG is equal to the 
angle BFG; therefore the 

base AG is equal^ to the base GB : and the circle described "^ **• 
from the centre G, at the distance GA, shall pass through the 
point B ; let this be the circle AHB : And because from the 
point A the extremity of the diameter AE AD is drawn at 

G 4 right 




88 



THE ELEMENTS 



^0 oiclii. right angles to AE, therefore AD*^ touches the circle j and be- 
'Cor. J6.5. cause AB drawn from the point 

of contact A cuts the circle, the 

angle DAB is equal to the angle 
«32, 3. in the alternate segment AriBs: 

But the angle D.\B is equal to 

the angle C, therefore also the 

angle C is equal to th.^ angle in 

the segment AHB : Wherefore, 

upon the given straight line AB 

the segment AHB of a circle is 

described, which contains an angle equal to the given angle 

at C. Which was to be done. 




PROP. XXXIV. PROB. 



» 17. 3. 



* 23. 1. 



e 32. 3. 



T 



O cut off a segment from a given 



shall contain an angle equal to a given 



circle which 
_ ^ ^^ rectilineal 

angle. 

Let ABC be the given circle, and D the given rectilineal 
angle ; it is required to cut off a segment from the circle ABC 
that shall contain an angle equal to the given angle D. 

Draw* the straight line EF touching the circle ABC in the 
point B, and at the p^int 
B, in the straight line BF, 
make'' the angle FBC 
equal to the angle D : 
Therefore, because the 
straight line EF touches 
the circle ABC, and BC is 
drawn from the point of 
contact B, the angle FBC 
is equal"^ to the angle in 
the alternate segment BAC 
of the circle; But the angle FBC is equal to the angle D:. 
therefore the angle in the segment BAC is equal to the angle 
D : Wherefore the segment BAC is cut off from the given 
circle ABC, containing an angle equal to the given angle D ; 
Which was to be done. 




OF EUCLID. 



PROP. XXXV. THEOR. 



89 



Uoox III. 




XF two straight lines within a circle cut one ano- 
ther, the rectangle contained hy the segments of 
one of them, is equal to the rectangle contained bj 
the segments of the other. 

Let the two straight lines AC, BD, within the circle 
ABCD, cut one another in the point E : the rechingle con- 
tained by AE, EC is equal to the redlangle 
contained by BE, ED. 

If AC, BD pass each of them through 
the centre, so that E is the centre ; it is 
evident, that AE, EC, BE, ED, being all 
equal, the redtangle AE, EC is likewise 
equal to the redangle BE, ED. 

But let one of them BD pass through the centre, and cut thfc 
other AC which does not pass through the centre, at right an- 
gles, in the point E : Then, if BD be bisected in F, F is the 
centre of the circle ABCD ; join AF : And because BD, which 
passes through the centre, cuts thcstraight line AC, which does 
not pass through the centre, at right 
angles in E, AE, EC are equal* to 
one another : And because the straight 
line BD is cut into two equal parts 
in the point F, and into two unequal 
in the point E, the rectangle BE, 
ED, together with the square of EF, 
is equal" to the square of FB ; that 
is, to the square of FA } but the squares 
of AE, EF axe equal ^ to the square 
of Fa ; therefore the reftangle BE, 
ED, together with the square of EF, 
is equal to the squares of AE, EF : Take away the common 
squareofEF, and the remaining rectangle BE, ED is equal to 
the remaining square of AE; that is, to the rectangle AE, EC, 

Next, let BD, which passes through the centre, cut the 
other AC, which does not pass through the centre, inE, but not 
at right angles : Then, as before, if BDbe bisected in F, F is 
fhe centre of the circle. Joio AF, and from F draw** FG per- 
pendicular 



Se«sr. 




»5.S. 



••5.1. 



47.1, 



* It. 1. 



fo 



THE ELEMENTS 



«47. 1. 




Book HI. pendicularto AC ; therefore AG is equal* to GC j wherefore 
VjT^s/ the re£bngle AE, EC, together with the square of EG is 
*5. 2. cquaP to the square of AG: To each of these equals add the 

square of GF j therefore the re£Vangle AE, EC, together with 

the squares of EG, GF is equal to 

the squares of AG, GF: But the 

squares of EG, GF are equal'^ to the 

square of EF ; "and the squares of 

AG, GF are e,qual to the square of 

AF : Therefore the rectangle AE, 

EC, together with the square of EF, 

is equal to the square of AF ; that 

is, to the square of FB : But the 

square of FB is equal^ to the reilangle BE, ED, together with 

the square of EF ; therefore the rectangle AE, EC, together 

with the square of EF, is equal to the redangle BE, ED, toge- 
ther with the square of EF : Take away the common square 

of EF, and the remaining redangle AE, EC, is therefore 

equal to the remaining rectangle BE, ED. ' 

Lastly, Let neither of the straight lines AC, BD pass 

through the centre ; Take the centre 

Sj and through E, the intersection of 

the straight lines AC, DB, draw 

the diameter GEFH : And because 

the redlangle AE, EC is equal, as 

has been shewn, 'to the reftangle GE, 

EH ; and, for the same reason, the 

rectangle BE, ED is equal to the 

same re£tangle GE, EH; therefore 

the re£tangle AE, EC is equal to the 




redangle BE, ED. 
Q,E. D. 



Wherefore, if two straight lines, &c. 



I 



PROP. XXXVl. THEOR. 



F from any point without a circle two straight 
lines be drawn, one of which cuts the circle, and 
the other touches it ; the rectangle contained by 
the whole line which cuts the circle, and the part 
of it without the circle, shall be equal to the 
square of the line which touches it. 

Let D be any point without the circle ABC, and DCA, DB 

two straight lines drawn from it,of which DCA cuts the circle, 

2 an4 



OF EUCLID. 



91 




18. 3. 



*>6.2. 



and DB touches the same : The reaangle AD, DC is equal ^^^^^ 
to the square of DB. 

Either DC A passes through the centre, or it does not ; first, 
let it pass through the centre E, and join EB j therefore the 
angle EBD is a right* angle : And be- 
cause the straight line AC is bise<5ied 
in E, and produced to the point D, the 
rectangle AD, DC, together with the 
square of EC, is equaP to the square of 
ED, and CE is ^qual to EB : Tbere- 
fore the reaangle AD, DC, together ij 
with the square of EB, is" equal to the 
square of ED: But the square of ED is 

equal'' to the squares of EB, BD, be- y / = 47. 1. 

cause EBD is a right angle : Therefore 
the reaangle AD^ DC, together with 
the square of £B, is equal to the squares 
of EB, BD: Take away the common 

square of EB ; therefore the remaining reftanglc AD, DC, is 
equal to the square of the tangent DB. 

But if DC A does not pass through the centre of the circle 
ABC, take "^ the centre E, and draw EF perpendicular^ to 

AC, anJjoinEB,EC,ED: And because the straight lineEF, 
which passes through the centre, cuts the straight line AC, 
which does not pass through the centre, 

at rigiit angles, it shall likewise bised^ y '3.3, 

it ; therefore AF is equal to FC : And 
because the straight line AC is bisefted 
in F, and produced to D, the reaangle 

AD, DC, together with the square of 
FC, is equal '' to the square of FD : 
To each of these equals add the square 
of FE ; therefore the reaangle AD, DC, 
together with the squares of CF, FE, 
is equal to the squares of DF, FE : But 
the square of ED is equal= to the squares 
of DF, FE, because EFD is a right an- 
gle : and the square of EC is equal to 
the squares of CF, FE ; therefore the reaangle AD, DC, to- 
gether with the square of EC, is equal to the square of ED: 
And CE is equal to EB ; therefore the reaangle AD, DC, to- 
gether with the square of EB, is equa.1 to the square of ED : 

But 



° 1.3. 
«r2. J. 




92 



THE ELEMENTS 



^^°^_^^2!J' ^"' '^^ squares of EB, BD are equal to the square"^ of ED, 
^"^^yi""^ because EBD is a right angle ; therefore the rectangles AD, 
DC, together with the square of EB, is equal to the squares 
of EB, BD ; Take away the common square of EB ; there- 
fore the remaining redtangle AD, DC is equal to the square 
of DB. Wherefore, if from any point, &c. Q. E. D. 

Cor. If from any point without a 
circle, there be drawn two straight lines 
cutting it, as AB, AC, the rectangles 
contained by the whole lines and the 
parts of them without the circle, are 
equal to one another, viz. the redlan- 
2;le BA, AE, to the rectangle CA, AF : 
For each of them is equal to the square 
ef the straight line AD which touches 
the circle. 




SeeN. 



« 17. 3. 
^ IB. 3. 

<• 36. 3. 



PROP. XXXVII. THEOR. 

XF from a point without a circle there be drawn 
two straight lines, one of which cuts the circle, and 
the other meets it; if the rectangle contained by the 
whole line which cuts the circle, and the part of it 
without the circle be equal to the square of the line 
>vhich meets it, the line which meets shall touch the 
circle. 

Let any point D be taken without the circle ABC, and 
from it let two straight lines DCAand DB be drawn, of which 
DCA cuts the circle, and DB meets it; if the reClangle AD, 
DC be equal to the square of DB ; DB touches the circle. 

Draw* the sifaight line DE, touching the circle ABC, find 
its centre F, and join FE, FB, FD ; then ^ ED is a right'' an- 
gle : And beeause DE touches the circle ABC, and DCA cuts 
it, the recStangle AD, DC is equa^ to the square of DE : But 
the rectangle AD, DC is, by hypothesis, equal to the square of 
DB : Therefore the square of DE is equal to the square of DB j 
and the straight line DE equal to the straight line DB : And 

fe; 



OF EUCLID. 



93 



!8. 1, 



FE is equal to FB, wherefore DE, EF are equal to DB, BF; ^^?^* 
and the base FD is common to the '' 

two triangles DEF, DBF ; therefore 
the angle DEF is equal** to the angle 
DBF; but DEF is a right angle, 
therefore also DBF is a right angle : 
And FB, if produced, is a diameter, 
and the straight line which is drawn 
at right angles to a diameter, from 
the extremity of it, touches' the cir- 
cle : Therefore DB touches the circle 
ABC. Wherefore, if from a point, 
&c. Q. E.-D. 




16>3. 



[ 94- ] 



THE 



ELEMENTS 



EUCLID. 




BOOK. IV. 

DEFINITIONS. 
I. 

r^^,^ jLx. Rectilineal figure is said to be inscribed in another 
SteNi redlilineal figure, when all the angles of the inscribed fi- 

gure are upon the sides of the figure in 
which it is inscribed, each upon each; 
II. 
In like manner, a figure is said to be described 
about another figure, when ail the sides of 
the circumscribed figure pass through the 
angular points of the figure about which it is described, 
each through each. 

III. 
A re£lilineal figure is said to be inscribed 
in a circle, when all the angles of the 
inscribed figure are upon the circumfe- 
rence of the circle. 

IV. 
A rectilineal figure is said to be described about 
when each side of the circumscribed figure 
touches the circumference of the circle. 
V. 
In like manner, a circle is said to be in- 
scribed in a redlilineal figure, when the 
circumference of the circle touches each 
side of the fiiiurc. 




THE ELEMENTS OF EUCLID 

VL 

A circle is said to be described about a rec- 
tilineal figure, when the circumference 
of the circle passes through all the angu- 
.lar points of the figure about which it is 
described. 

VIL 

A straight line is said to be placed in a cir- 
cle, when the extremities of it are in the circumference of 
the circle. 




PROP. L PROS. 



In a given circle to place a straight line, equal to 
a given straight line not greater "than the diameter 
of the circle. 

Let ABC be the given circle, and D the given straight line, 
not greater than the diameter of the circle. 

Draw BC the diameter of the circle ABC j then, if BC is 
equal to D, the thing required is done ; for in the circle ABC 
a straight line BC is placed 
equal to D : But, if it is not, 
BC is greater than D j make 
CE equal* to D, and from the 
centre C, at the distance CE, 
describe the circle AEF, and 
join CA: Therefore, because 
C is the centre of the circle 
AEF, CA is equal to CE ; 
but D is equal to CE ; therefore D is equal to CA: Where- 
fore in the circle ABC, a straight line is placed equal to the 
given straight line D, which is not greater than the diameter 
•f the wrcle. Which was to be done. 




PROP. U. PROB. 



IX'a given circle to inscribe a triangle equiangu- 
lar to a given triangle. 

i-et 




» 17. 3. 

*23. J. 



« 32. 3. 



'»32. 1. 



THE ELEMENTS 

Let ABC be the given circle, and DEF the given trian- 
gle ; it is required to inscribe in the circle ABC a triangle 
equiangular to the triangle DEF. 

Dravi'^ the straight line GAH touching the circle in the 
point A, and at the point A, in the straight line AH, make" 
the angle HAC equal to the angle DEF ; and at the point A 
in the straight line AG, 
n?ake the angle GAB 
equal to the angle DFE, 
and join BC : There- 
ibre, because HAG 
touches the circle ABC, 
and AC is drawn from 
the point of contact, the 
angle HAC is equal'^ 
to the angle ABC in the 
alternate segment of the circle: But HAC is equal to the an- 
gle DEF : therefore also the angle ABC is equal to DEF : 
For the same reason, the angle ACB is equal to the angle 
DFE ; therefore the remaining angle BAC is equaF to the 
remaining angle EDF : Wherefore the triangle ABC is 
equiangular to the triangle DEF, and it is inscribed in the 
circle ABC. Which was to be done. 




PROP. HI. PROB. 



»«3. I. 



17.3. 



<^ 18, 3. 



About a given circle to describe 
equiangular to a given triangle. 



triangle 



Let ABC be the given circle, and DEF the given triangle; 
it is required to describe a triangle about the circle ABC 
equiangular to the triangle DEF. 

Produce EF both ways to the points, G, H, and jRnd the 
centre K of the circle ABC, and from it draw any straight line 
KB ; at the point K, in the straight line KB, make* the angle 
BKA equal to the angle DEG, and the angle BKC equal to the 
angle DP'H ; and through the points A, B, C, draw the 
straight lines LAM,"MBN, NCL, touching'' the circle ABC : 
Therefore, because LM, MN, NL touch the circle ABC in the 
points A, B, C, to which from the centre are drawn KA, KB, 
KC, the angles at the points A, B, G, are righf^ angles : And 
because the four angles of the quadrilateral figure AMBK are 

equal 



OF EUCLID. 



97 



Kqual to four right angles, for it can be divided into two trian- Book IY. 
gles ; and that two of them KAM, KBM are right angles, the '^•^''^^''*^^ 
other two AK.B, 
AMB are equal to 
two right angles : 
But the angles 
DEG, DEF are 
likewise equal" to 
two right angles ; 
therefore the an- 
gles AKB, AMB 
are equal to the 
angles DEG, 
DEF, of which 
AKB is equal to DEG ; wherefore the remaining angle'AMB 
is e'qual to the remaining angle DEF : In like manner, the an- 
gle LNM may be demonstrated to be equal to DFE ; and 
therefore the remaining angle MEN is equal'' to the remain- e 22. i, 
ing angle EDF: Wherefore the triangle LMN is equiangu- 
lar to the triangle DEF : And it is described about the circle 
ABC. Which was to be done. 




PROP. IV. PROB. 
O inscribe a circle in a siven trian ole. 



Se«N. 



Let the given triangle be ABC j it is required to inscribe a 
circle in ABC, 

Bised* the angles ABC, BCA by the straight lines BD, CD ^ ^ ^ 
meeting one another in the point D, from which draw** DE, ^ ^3. 1. 
DF, DG perpendiculars to AB, 

BC, CA : And because the angle 
EBD is equal to the angle FBD, 
for the angle ABC is biseded by 

BD, and that the right angle 
BED is equal to the right angle 
BFD, the two triangles EB'D, 
FBD have two angles of the one 
equal to two angles of the other, 
^and the side BD, v/hich is oppo- 
site to one of the equal angles in 
each, is common to both ; there- 
fore their ether sides shall b» 

H equal ; 




^8 



Book IV. 



26. 1. 



"16.3. 



THE ELEMENTS 

equa^ } wherefore DE is equal to DF : For the same reason, 
DG is equal to DF ; therefore the three straight linesDE, DF, 
DG are equal to one another, and the circle described from the 
centre D, at the distance of any of them, shall pass through 
the extremities of the other two, and touch the straight lines 
AB, BC, CA, because the angles at the points E, F, G, are 
right angles, and the straight line which is drawn from the 
extremity of a diameter at right angles to.it, touches'* the cir- 
cle: Therefore the straight lines AB, BC, CA do each of 
them touch the circle, and the circle EFG is inscribed in the 
triane;le ABC. Which was to be done. 



PROP. V. PROB. 

SecN ' j[ o describe a circle about a given triangle. 

Let the given triangle be ABC j it is required to describe 
a circle about ABC. 
, )o_ 1 Biseft* AB, AC in the points D, E, and from these points 

» 11. u draw DP^,EF at right angles'' to AB, AC ; DF, EF produced 





C B 




«*. 1. 



meet one another : For, if they do not meet, they are parallel 
wherefore AB, AC, which are at right angles to them, are pa 
rallel ; which is absurd : Let them meet in F, and join FA 
also if the point F be not in BC, join BF, CF : Then, bccaus( . 
AD is equal to DB, and DF common, and at right angles t( 
AB, the base AF is equal*= to the base FB . In like manner, i 
may be shown that CF is equal to FA ; and therefore BF f 
equal to FC i and J'A, FB, FC are equal to one anothef 

wherefor 



OF EUCLID. 



99 



wherefore the circle described from the centre F, at the dis- ^"o"^ ^• 
tance of one of them, shall pass through the extremities of the ^"^ 
other two, and be described about the triangle ABC. Which 
was to be done. 

Cor. And it is manifest, that when the centre of the circle 
falls within the triangle, each of its angles is less than a right 
angle, each of them being in a segment greater than a semicir- 
cle j but, when the centre is in one of the sides of the trian- 
gle, the angle opposite to this side, being in a semicircle, is a 
right angle ; and, if the centre falls without the triangle, the 
angle opposite to the side beyond which it is, being in a seg- 
ment less than a semicircle, is greater than a right angle : 
Wherefore, if the given triangle be acute angled, the centre 
of the circle falls within it ; if it be a right angled triangle, 
the centre is in the side opposite to the right a- .gle ; and, if it 
be an obtuse angled triangle, the centre fails without the tri- 
angle, beyond the side opposite to the obtuse angle. 



PROP. VI. PROB. 

1 O inscribe a square in a given circle. 

Let ABCD be the given circle j it is required to inscribe a 
square in ABCD. 

Draw the diameters AC, BD, at right angles to one another; 
.and join AB, BC, CD, DA; because BE is equal to ED, for 
E is the centre, and that EA-is com- 
mon, and at right angles to BD ; the 
base BA is equal* to the base AD ; 
and, for the same reason, BC, CD 
are each of them equal to BA, or 
AD ; therefore the quadrilateral fi- 
gure ABCD is equilateral. It is al- 
o rectangular ; for the straight line 
;D, being the diameter of the circle 
iiBCD, BAD is a semicircle ; where- 
fore the angle BAD is a right'' an- 
te i for the same reason each of the angles ABC, BCD, CDA, 
a right angle; therefore the quadrilateral figure ABCi^ is 
rectangular, and it has been shown to be equilateral; there- 
fore it IS a square ; and it is inscribed in the circle ABCD 
' v'hich was to be done. 




4. I, 



'31. 



I 



H2 




• Its. 3. 



'> 1». 3. 



e.28.1. 



« 54. 1. 



THE ELEMENTS 



PROP. VII. PROB. 



X O describe a square about a given circle. 

Let ABCD be the given circle j it is required to describe a 
square about it. 

Draw two diameters AC, BD of the circle ABCD, at right 
angles to one another, and through the points A, B, C, D, 
draw* FG, GH, HK, KF touching the circle ; and because 
FG touches the circle ABCD, and EA is drawn from the centre 
E to the point of contadl A, the angles at A are right'' angles ; 
for the same reason, the angles at the points B, C, D are right 
angles ; and because the angle AER is 
a right angle, as liicewise is EBG, GH 
is parallel"^ to AC ; for the same rea- 
son, AC is parallel to FK, and in like 
manner GF, HK may each of them be 
demonstrated to be parallel to BED ; 
therefore the figures GK, GC, AK, 
FB, BK are parallelograms ; andGF 
is therefore equaH to HK, and GH 
to ¥K ; and because AC is equal to 
BD, and that AC is equal to each of the two GH, FK ; and BD 
to each of the two GF, HK ; GH, FK are each of them equal 
to GF or HK ; therefore the quadrilateral figure FGHK i> 
equilateral. It is also rectangular ; for GBEA being a parallel- 
ogram, and AEB a right angle, AGS'* is liicewise a right an- 
gle: In the same manner it may be shown that the angles at 
H, K, F are right angles j therefore the quadrilateral figure 
FGHK is rectangular, and it was demonstrated to be equila- 
teral ; therefore it is a square ; and it is described about the 
ckcle ABCD, Which was to be done. 



B 



a A F 


E 


1 


s 


y 


a c K 



»10. 1. 
••31.1. 



PROP. Vin. PROB. 

X O inscribe a circle in a given square. 

Let ABCD be the given square ; it is required to inscribe 
a circle in ABCD. 

Bisect* each of the sides AB, AD, in the points F, E, and 
through E draw ^ EH parallel to AB or DC, and through F 

drav( 



I 



OF EUCLID. 101 

draw FK parallel to AD or BC; therefore each of the figures AK, BookIV. 
KB, AH, HD, AG, GC, BG, GD, is a parallelogram, and ""p"^*^ 
their opposite sides are equal''; and because AD is equal to AB, ^ "^■*' ^• 
and that AE is the half of AD, and' AF the half of AB, AE 
is equal to AF ; wherefore the sides 

opposite to these are equal, viz. FG to A E D 

GE ; in the same manner it may be de- 
monstrated that GH, GK. are each of 
them equal to FG or GE : therefore 

the four straight lines GE, GF, GH, F| ^ JK. 

GK are equal to one another; and the 

circle described from the centre G at 

the distance of one of them, shall pass _ _ 

through the extremities of the other B H 

three, and touch the straight lines AB, 

BC, CD, DA; because the angles at the points E, F, H, K, are 

right "^ angles, and that the scraight line which is drawn from "29.1. 

the extremity of a diameter, at right angles to it, touches the 

circle*^; therefore each of the straight lines, AB, BC, CD, '^^•^• 

DA touches the circle, which therefore is inscribed in the 

square ABCD. Which was to be done. 

PROP. IX. PROB. 



X O describe a circle about a given square. 

Let ABCD be the given square j it is required to describe 
a circle about it. 

Join AC, BD, cutting one another in E ; and because DA is 
•qual to AB, and AC common to the triangles DAC, BAG, 
the two sides DA, AC are equal to the 
two BA, AC, and the base DC is equal 
to the base BC ; wherefore the angle 
DAC is equal^ to the angle BAC, and 
the angle DAB is bisected by the straight 
line AC : In the same manner, it may be 
demonstrated that the angles ABC, BCD, 
CDA are severally bisected bythe straight 
lines BD, AC ; therefore, because the 
angle DAB is equal to the angle ABC, and that the an^le 
EAB is the half of DAB, and EBA the half of ABC: The 
angle EAB is equal to the angle EBA ; wherefore the side 
JEA is equal'' to the side EB : In the same manner, it may be *^' ^* 

H 3 d«monstratait. 




8.1. 



io2 



THE ELEMENTS 



Book IV. demonstrated, that the straight lines EC, ED are each of 
^'*^^''*^ them equal to EA or EB ; therefore the four straight lines 
EA, EB, EC, ED, are equal to one another ; and the circle 
described from the centre E, at the distance of one of them, 
shall pass through the extremities of the other three, and be 
described about the square ABCD, Which was to be done. 



»11. 2. 



*1. 4. 
<: 5. 4. 



« 37. 3. 



• »«. 3. 



''32.1. 



PROP. X. PROB. 

A O describe an isosceles triangle, having each of 
the angles at the base double of the third angle. 

Take any straight line AB, and divide* it in the point C, so 
that the redlangle AB, BC be equal to the square of CA ; and 
from the centre A, at the distance AB, describe the circle BDE, 
in which place'' the straight line BD equal to AC, which is not 
greater than the diameter of the circle BDE j join DA, DC, 
and about the triangle ADC describe^ the circle ACD j the 
triangle ABD is such as is required, that is, each of the angles 
ABD, ADB is double of the angle BAD. . . . 

Because the re6langle AB, BC is equal to the square of AC, 
and that AC is equal to BD, the reilangle AB, BC is equal to 
the square of BD ; and because 
from the point B, without the 
circle ACD, two straight lines 
BCA, BD are drawn to the cir- 
cumference, one of which cuts, 
and the other meets the circle, 
and that the rectangle AB, BC, 
contained by the whole of the 
cutting line, and the part of it 
without the circle, is equal to the 
square of BD which meets it ; 
the straight line BD touches^ 
the circle ACD ; and because 
BD touches the circle, and DC 
is drawn from the point of con- 
ta€t D, the angle BDOis equal«= to the angle DAC in the 
alternate segment of the circle ; to each of these add the angle 
CDA ; therefore the whole angle BDA is equal to the two 
angles CDA, DAC ; but the exterior angle BCD is equaF to 
the angle CDA, DAC j therefore also BDA is equal to BCD ; 

but 




OF EUCLID. 



103 



but BDA is equal? to the angle CBD,because the side AD ^°°^ ^^' 
is equal to the side AB ; therefore CBD, or DBA", is equal to g^"^Y^ 
BCD ; and consequently the three angles BDA, DBA, BCD, 
are equal to one another ; and because the angle DBC is equal 
to the angle BCD, the side BD is equaP to the side DCj but ''^- ^• 
BD was made equal to CA, therefore also CA is equal to 
CD, and the angle CDA-equal? to the angle D AC ; there- 
fore the angles CDA, DAC together, are double of the angle 
DAC : But BCD is equal to the angles CDA, DAC ; there- 
fore also BCD is double of DAC, and BCD is equal to each 
of the angles BDA, DBA j each therefore of the angles BDA, 
DBA is double of the angle DAB wherefore an isosceles tri- 
angle ABD is described, having each of the angles at the base 
double of the third an^le. Which was to be done. 



PROP XI. PROB. 



X O inscribe an equilateral and et|uiangular pen- 
tagon in a given circle. 

Let ABCDE be the given circle; it is required to inscribe 
an equilateral and equiangular pentagon in the circle ABCDE. 

Describe^an isosceles triangle FGH, having each of the an- * ^^- *• 
gles at G, H, double ot the angle at F j and in the circle 
ABCDE inscribe'' the triangle ACD equiangular to the trian- '' 2. *• 
gle FGH, so that the angle 

CAD be equal to the angle j^ 

at F, and each of the angles 
ACD, CDA equal to the 
angle at G or H, wherefore 
each of the angles ACD, 
CDA is double of the angle 
CAD. Bisect'^ the angles 
ACD, CDA by the straight 
lines CE, DB j and join AB, 
BC, DE, EA. ABCDE 
is the pentagon required. 

Because each of the angles ACD, CDA is double of CAD, 
and are bisected by the straight lines CE, DB, the'five angles 
DAC, ACE, ECD, CDB, BDA are equal to one another ; 
but equal angles stand upon equal"* circumferences ;' therefore « 26. ». 
the five circumferences A B, BC, CD, DE, EA are equal to one 

H4 another: 




9. 1. 



104 * THE ELEMENTS 

Book IV. another : And equal circumferences are subtended by equal' 
^^^3^ Straight lines J therefore the five straight lines AB, BC, CD, 
DE, EA are equal to one another. Wherefore the pentagon 
ABCDE is equilateral. It is also equiangular j because the 
circumference AB is equal to the circumference DE: If to- 
each be added BCD, the whole A BCD is equal to the whole 
EDCB : And the angle AED stands on the circumference 
ABCD, and the angle BAE on the circumference EDCB; 
f2^, 3. therefore the angle BAE is equaF to the angle AED : For the 
same reason, each of the angles ABC, BCD, CDE is equal to 
the angle BAE, or AED: Therefore the pentagon ABCDE 
is equiangular ; and it has been shown that it is equilateral. 
Wheretore, in the given circle, an equilateral and equiangular 
pentagon has been inscribed. Which was to be done. 



PROP. XII. PROB. 

X O tlescribe an equilateral and equiangular pen- 
tagon about a given circle. 

Let ABCDE be the given circle ; it is required to describe 
an equilateral and equiangular pentagon about the circle 
ABCDE. 

Let the angles of a pentagon, inscribed in the circle, by the 
last proposition, be in the points A, B, C, D, E,sothat thecir- 

•^ n. 4. cumferences AB, BC, CD, DE, EA are equal'' ; and throujjh 
the points A, B, C, D, E, draw GH, HK, KL, LM, Ut}, 

* 17. 3. touching'' the circle ; take the centre F, and join FB, FK, FC, 
FL, FD : And because the straight line KL touches the circle 
ABCDE in the point C, to which FC is drawn from the ccn- 

f 18. 3. tre F, FC is perpendicular"" to KL ; therefore each of the an- 
gles at C is a right angle : For the same reason, the angles at 
the points B, D are right angles : And because FCK is a right 

"47. 1. angle, the square of FK is equal"" to the squares of FC, CK ; 
For the same reason, the square of FK is equal to the squares 
of FB, BK : Therefore the squares of FC, CK are equal to the 
squares of FB, BK, of which the square of FC is equal to the 
square gf FB ; the remaining squar* of CK is therefore equal to 

the 



OF EUCLID. 105 

the remaining square of BK, and the straight line CK equal to BookIV, 
BK : And because FB is equal to FC, and FK common to the ^"^^''^'^ 
triangles BFK,CFK,thetwoBF,FKare equal tothetwoCF, 
FK j'and the base BK is equal to the base KC ; therefore the 
angle BFK is equal= to the angle KFC, and the angle BKF to -^S. I. 
FKC i wherefore the angle BFC is double of the angle KFC, 
and BK.C double of FKC": For the same reason, the angle CFD 
is double of the angle CFL, and CLD double of CLF : And be- 
cause the circumference BC is equal to the circumference CD, 
the angle BFC is equaF to the ^ f 27. 3. 

angle CFD ; and BFC is dou- 
ble of the an^le KFC, and 
CFD double ot CFL ; there- 
fore the angle Kr C, is equal to 
the angle CFL : and the right 
angle FCK is equal to the right 
angle FCL : Therefore, in the 
twotrianglesFKCjFLC, there 
are two angles of one equal to 
two angles of the other, each 
toeach, and the side FC, which 

is adjacent to the equal angles in each, is common to both; 
therefore the other sides shall be equals to the other sides, and ^26. 1. 
the third angle to the third angle : fherefore the straight line 
KC is equal to CL, and the angle FKC to the angle FLC : 
And because KC is equal to CL, KL is double of KC : In the 
same manner it may be shown that HK is double of BK : And 
because BK is equal to KC, as was demonstrated, and that KL 
is double of KC, and HK double of BK, HK shall be equal to 
KL : In like manner, it may be shown that GH, GM, ML 
are each of them equal to HI^ orKL: Therefore the pentagon 
GHKLM is equilateral. It is also equiangular ; for, since the 
angle FKC is equal to the angle FLC, and that the angle 
UKLis double oftheangleFKC, and KLMdoubleof FLC, as 
was before demonstrated, the angle HKL is equal to KLM: 
And in like manner it may be shown, that each of the angles 
KHG, HGM, GML is equal to the angle HKL or KLM : 
Therefore the five angles GHK, HKL, KLM, LMG MGH 
being equal to one another, the pentagon GHKLM is equian- 
gular : And it is equilateral, as was demonstrated j and it is de« 
jcribed about the circle ABCDE. Which was to be done. 





•9.1. 



*4. 1. 



•!K.l. 



THE ELEMENTS 

PROP. XIII. PROB. 

1 O inscribe a circle in a given equilateral and 
equiangular pentagon. 

Let ABCDE be the given equilateral and equiangular penta- 
gon ; it is required to inscribe a circle in the pentagon 
ABCDE. 

Bisect* the angles BCD, CDE by the straight lines CF DF, 
and from the point F, in which they meet, drav/ the straight 
lines FB, FA, FE : Therefore since BC is equal to CD, and 
CF common to the triangles BCF, DCF the two sides BC, CF 
are equal to the two DC,CF ; and the angle BCF is equal to 
the angle DCF ; therefore the base BF is equal'' to the base FD, 
and the other angles to the other angles, to which the equal sides 
are opposite ; therefore the angle CBF is equal to the angle 
CDF : And because the angle CDE is double of CDF, and 
that CDE is equal to CBA, and 
CDF to CBF; CBA is also 
double of the angle CBF; there- 
fore the angle ABF is equal to 
the angle CBF ; wherefore the 
angle ABC is bisected by the 
straight line BF ; In the same 
manner it may be demonstrated, 
that the angles BAE, AED, 
are bisected by the straight lines 
AF, FE: From the point F 
draw^ FG, FH, FK, FL, FM 
perpendiculars to the straight 
lines AB, BC, CD, DE, EA : 

And because the angle HCF is equal to KCP, and the right 
angle FHC equal to the right angle FKC ; in the triangles 
FHC, FKC there are two angles of one equal to two angles 
of the other, and the side FC, which is opposite to one of 
the equal angles in each, is common to both ; therefore the 
other sides shall be equal^, each to each; wherefore the per- 
pendicular FH is equal to the perpendicular FK : In the 
same manner it may be demonstrated ; that FL, FM, FG are 
each of them equal to FH or FK : Therefore the five straight 
lines FG, FH, FK, FL, FM are equal to one another : ' 
Wherefore the circle described from the centre F, at the dis- 
tance of one of these five, shall pass through the extremities of 

2 the.' 




OF EUCLID. 



107 



the other four, and touch the straight lines AB, BC, CD, DE, Book iv, 
EA, because the angles at the points G, H, K, L, M are ^"^^^^^ 
right angles ; and that a straight line drawn from the extre- 
mity of the diameter of a circle at right angles to it, touches^ * 16. 3, 
the circle : Therefor*each of the straight lines AB, BC, CD, 
DE, EA touches the Circle : wherefore it is inscribed in the 
pentagon ABCDE. Which was to be done. 



PROP. XIV. PROB. 



A O describe a circle about a given equilateral and 
equiangular pentagon. 

Let ABCDE be the given equilateral and equiangular pen- 
tagon : it is required to describe a circle about it. 

Bisea=' the angles BCD, CDE by the straight lines CF, FD, • 9. 1. 
and from the point F, in which they meet, draw the straight 
iines FB^ FA, FE to the points B, 
A, E. It may be demonstrated, in 
the same manner as in the preceding 
proposition, that the angles CBA, 
BAE, AED are biseded by the 
straight lines FB, FA, FE: And 
because the angle BCD is equal to . 
the angle CDE, and that FCD is 
the half of the angle BCD, and CDF, 
the half of CBE ; the angle FCD is 
equal to FDC ; wherefore the side 
CF is equaP to the side FD : In like manner it may be de- " 5. i 
monstrated that FB, FA, FE, are each of them equal to FC 
or FD : Therefore the five straight lines FA, FB, FC, FD, 
FE are equal to one another j and the circle described from 
the centre F, at the distance of one of them, shall pass through 
the extremities of the other four, and be described about the 
equilateral and equiangular pentagon ABCDE. Which was 
to be done. 





THE ELEMENTS 
PROP. XV. PROB. 

seeN. X O inscribe an equilateral and equiangular hexa- 
gon in a given circle. 

Let ABCDEF be the given circle j it is required to inscribe 
an equilateral and equiangular hexagon in it. 

Find the centre G of the circle ABCDEF, and draw the di- 
ameter AGD J and from D as a centre, at the distance DG, 
describe the circle FGCH, join EG, CG, and produce them 
to the points B, F ; and join AB, BC, CD, DE, EF, FA : 
The hexagon ABCDEF is equilateral and equiangular. 

Because G is the centreof the circle ABCDEF, GE is equal 
to GD : And because D is the centre of the circle EGCH, DE 
is equal to DG ; wherefore G£ is equal to ED, and the tri- 
angle FGD is equilateral; and therefore its three angles EGD, 
GDE, DEG, are equal to one another, because the angles at 
the base of an isosceles triangle are equaP; and the three angles 
of a triangle are equal'' to two right angles ; therefore th» 
angle EGD is the third part of two right angles; In the same 
manner it may be demonstrated, that 
the angle DGC is also the third part 
of two right angles : And because the 
straight line GC makes with EB the 
adjacent angles EGC, CGB equaP 
to two right angles; the remaining 
angle CGB is the third part of two 
right angles ; therefore the angles 
EGD, DGC, CGB are equal to one 
another : And to these are equal'' the 
vertical opposite angles BGA, AGF, 
FGE : I'herefore the six angles EGD, 
DGC, CGB, BGA, AGF, FGE, 
are equal to one another : But equal 
angles stand upon equal'= circumfe- 
rences ; therefore the six circumfe- 
rences AB, BC, CD, DE, EF, FA are equal to one another . 
And equal circumferences are subtended by equal*" straight 
lines ; therefore the six straight lines are equal to one another, 
and the hexagon ABCDEF is equilateral. It is also equiangu- 
lar : for, since the circumference AF is equal to ED, to each of | 
these add the circumference ABCD ; therefore the whole cir- 
cumference FABCD shall be equal to the whole EDCBA : 

An«i 



»5. 1. 
»32. i. 



"=13. 1. 



* 15. 1. 



26.3. 



^29.3. 




OF EUCLID 



109 



And the angle FED stands upon the circumference FABCD, Book rv. 
and the angle AF£ upon EDCBA ; therefore the angle AF£ ^■^''''*^. 
is equal to FED : In the same manner it mav he demonstrated 
that the other angles of the hexagon ABCDEF are each of 
them equal to the angle AFE or FED : Therefore the hexa« 
gon is equiangular; and it is equilateral, as was shown; and it 
is inscribed in the given circle ABCDEF. Which was to be 
done. 

Cor. From this it is manifest, that the side of the hexajon 
is equal to the straight line from the centre, that is, to the 
semidiameter of the circle. 

And if through the points A, B, C, D, E, F there be drawn 
straight lines touching the circle, an equilateral and equiangu- 
lar hexagon shall be described about it, which may be demon- 
strated from what has been said of the pentagon ; and likewise 
a circle may be inscribed in a given equilateral and equiangu- 
lar hexagon, and circumscribed about it, by a method like to 
that used for the pentagon. 

PROP. XVI. PROB. 

L O inscribe an equilateral and equiangular quin- seeN. 
decagon in a given circle. 

Let ABCD be the given circle ; it is required to inscribe an 
equilateral and equiangular quindecagon in the circle ABCD. 

Let AC be the side of an equilateral triangle inscribed* in ,2 4 
the circle, and AB the side of an equilateral and equiangular 
pentagon inscribed'' in the same ; therefore, of such equal parts h u 4 
as the whole circumference ABCDF contains fifteen, the cir- 
cumference ABC, being the third 
part of the whole, contains five; and 
the circumference AB, which is the 
fifth part of the whole, contains 
three ; therefore BC their difference 
contains tv/o of the same parts : Bi- 

aea^BCinE; therefore BE, EC ""W V/ ^30.3. 

are, each of them, the fifteenth part 
©f the whole circumference ABCD: 
Therefore, if the straight lines BE, 

EC be drawn, and straight lines equal to them- be placed 
round'' m the whole circle, an equilateral and equiangular qum- d 1 4 
<iecagon shall be inscribed in it. Which was to be done. 

And, 




no 

Book IV. 



THE ELEMENTS OF EUCLID. 

And, in the same manner as was done in the pentagon, if, 
through the points of division made by inscribing the quinde- 
cagon, straight lines be drawn touching the circle, an equila- 
teral and equiangular quindecagon shall be described about it; 
And likewise, as in the pentagon, a circle may be inscribed in 
a given equilateral and equiangular quindecagon, and circum- 
scribed about it. ' 



A 



[ iii ] 
THE 

E L E AI E X T S 

OF 

EUCLID. 

BOOK V. 

DEFINITIONS. 

I. 



LESS magnitude is said to be a part of a greater mag- Book V. 
nitude, ,^hen the less measures the greater y that is, v^»v-^/ 
'when the less is contained a certain number of times ex- 
' actly in the greater.' 

A greater magnitude is said to be a multiple of a less, when the 
greater is measured by the less ; that is, ' when the greater 
* contains the less a certain numl>er of times exactly.* 
III. 
* Ratio is a mutual relation of two magnitudes of the same See N. 
kind to one another, in respecl of quantity.' 
IV. 

Magnitudes are said to have a ratio to one another, when the 
less can be multiplied so as to exceed the other. 

V. 
The first of four magnitudes is said to have the same ratio to 
the second, which the third has to the fourth, when any 
equimultiples whatsoever ef the first and third being taken, 
and any equimultiples whatsoever of the second and fourth ; 
if the multiple of the first be less than that of the second, 
the multiple of the third is also less than that of the fourth ; 
or, if the multiple of the first be equal to that of the second, 
the multiple of the third is also equal to that of the fourth ; 

or. 



Hi THE ELEME^fTS. 

Book V. orj if the multiple of the first be greater than that of the se- 
'^■^'^■'^^^ cond, the multiple of the third is also greater than that of 
the fourth. 

VI. 
Magnitudes which have the same ratio are called proportion- 
als. N. B. ' When four magnitudes are proportionals, it is 

* usually exprest by saying, the first is to the second, as the 

* third to the fourth.' 

VII. 
When of the equimultiples of four magnitudes (taken as in 
the fifth definition), the multiple of the first is greater than 
that of the second, but the multiple of the third is not 
greater than the multiple of the fourth ; then the first is 
said to have to the second a greater ratio than the third 
magnitude has to the fourth ; and, on the contrary, . the 
third is said to have to the fourth a less ratio than the first 
has to the second. 

VIII. 
AnalogVj or proportion, is the similitude of ratios. 

Proportion consists in three terms at least. 

X. 
See ^t When three magnitudes are proportionals, the first is said to 
have to the third the duplicate ratio of that which it has to 
the second. 

XI. 

When four magnitudes are continual proportionals, the first is 

said to have to the fourth the triplicate ratio of that which it 

has to the second, and so on, qnadruplicate, See. increasing 

the denomination still by unity, in any number of propor- 

' tionals. 

Definition A, to wit, of compound ratio. 
When there are any number of magnitudes of the same kind, 
the first is said to have to the last of them the ratio com- 
pounded of the ratio which the first has to the second, and 
of the ratio which the second has to the third, and pf the 
ratio which the third 'has to the fourth, and soon unto the 
last magnitude. 
For example, if A, B, C, D, be four magnitudes of the sam 
kind, the first A is said to have to the last D the ratio com-! 
pounded of the ratio of A to,B, and the ratio of B to C, 
and of the ratio C to D, or, the ratio of A to D is said to 
be compounded of the ratios of A to B, B to C, and C to D : 

And 



1 



' OF EUCLID. n3 

And if A has to B the same ratio which E has to F ; and I^ Book \ . 
to C, the same ratio that G has to H ; and C to D, the same '""^^•'^^ 
that K has to L ; then, by this definition, A is said to have 
to D the ratid compounded of ratios which are the same with 
the ratios of E to F, G to H, and K to L : And the same 
thing is to be understood when it is more briefly expressed, 
by saying, A has to D the ratio compounded of the ratios 
of E to F, G to H, and K to L. 

In like manner, the same things being supposed, if M has to N 
the same ratio which A has to D j then, tor shortness sake, 
M is said to have to N, the ratia compounded of the ratios 
of E to F, G to H, and K to L. 
XII. 

In proportionals, the antecedent terms are called homologous 
to one another, as also the consequents to one another. 

* Geometers make use of the following technical words to sig- 
' nify certain ways of changing either the order or magni- 
' tude of proportionals, so as that they continue still to be 
* proportionals.' 

XIII. 

Permutando, or alternando, by permutation, or alternately. 
This word is used when there are four proportionals, and it s« N. 
is inferred, that the first has the same ratio to the third, 
which the second has to the fourth ; or that the first is to 
the third, as the second to the fourth : As is shewn in the 
1 6th prop, of this 5th book. 

XIV. 

Invertendo, by inversion : When there are four proportionals, 
and it is inferred, that the second is to the first, as the fourth 
to the third. Prop. B. Book 5. 
XV. 

.Componendo, by composition j when there are four propor- 
tionals, and it is inferred, that the first, together with the 
second, is to the second, as the third, together with the 
fourth, is to the fourth. i8th Prop. Book 5. 
XVI. 

Dividendo, by division j when there are four proportionals, and 
it is inferred, that the excess of the first a&ve the second, 
is to the second, as the excess of the third above the fourth, 
is to the fourth. 17th Prop. Book 5. 
XVII. 

Convertendo, by conversion ; when there are four proportion- 
als, and it is inferred, that the first is to its excess above the 

I second, 



114 THE ELEMENTS 

SooK y. second, as the third to its excess above the fourth. Prop. 
"^-^^^^ E. Book 5. . • 

XVIIl. 

Ex oequali (sc. distantia)^ or ex sequo, from equality of dis- 
tance i when there is any number of magnitudes more than 
two, and as many others, so that they are proportionals when 
taken two and two of each rank, and it is inferred, that the 
first is to the last of the first rank of magnitudes, as the first 
is to the last of the others : * Of this there are the two fol- 

* lowing kinds, which arise from the different order in which 

* the magnitudes are taken, two and two.' 

XIX. 
■ Ex oequali, from equality. This term is used simply by Itself, 
when the first magnitude is to the second of the first rank, 
as the first to the second of the other rank ; and as the second 
is to the third of the first rank, so is the second to the third 
of the other j and so on in order, and the inference is as men- 
tioned in the preceding definition ; whence this is called or- 
dinate proportion. It is demonstrated in 22nd Prop. Book 5. 
XX. 
Ex aequali, in proportione perturbata, seu inordinata, from 
equality, in perturbate or disorderly proportion*. This 
term is used when the first magnitude is to the second of 
the first rank, as the last but one is to the last of the second^ 
rank ; and ^s the second is to the third of the first rank, so 
is the last but two to the last but one of the second rank 
and as the third is to the fourth of the first rank, so is th< 
third from the last to the last but two of the Second rank -3 
and so on in a cross order : And the inference is as in tl 
1 8th definition. It is demonstrated in the 23d -Prop. 
Book 5. 



AXIOMS. 

I. 



JtLqyiMtJLTiPLES of the same, or of equal magnitudes, arwi 
equal to one another. ^f. 

II. Those' 

* 4 Prop, lib, 2. Archimedis dc tphira et cjlindro. 



OF EUCLID. 115 

IJ, Book V. 

lose magnitudes of which the same, or equal magnitudes, '^^^"^ 
are equimultiples, are equal to one another. 

III. 

multiple of a greater magnitude is greater than the same 
multiple of a less. 

IV. 

hat magnitude of which a multiple is greater than the same 
multiple of another, is greater than that other magnitude. 

PROP. I. THEOR. 



G 



H 



D 



F any number of magnitudes be equimultiples of 
many, each of each ; what multiple soever any 
e of them is of its part, the same multiple shall all 
; first magnitudes be of all the other. 

et any number of magnitudes AB, CD be equimultiples 
AS many others E, F, each of each ; whatsoever multiple 
is of E, the same multiple shall AB and CD together be 
and F together, 
lecause AB is the same multiple of E that CD is of F, as 
y magnitudes as are in AB equal to E, so many are there 
D equal to F. Divide AB into magnitudes 
il to E, viz. AG, GB ; and CD into CH, 
equal each of them to F : The number 
jfore of the magnitudes CH, HD shall be 
I to the number of the others AG, GB : 
because AG is equal to E, and CH to F, 

jfore AG and CH together are equal to» E B ' Ax. 2. 5. 

F together : F«r th^ same reason, because 
's equal to E, and HD to Fj GB and HD 
her are equal to E and F together. Where- 
as many magnitudes as are in AB equal 
so many are there in AB, CD together 
to E and F together. Therefore, what- 
r multiple AB is of E, the same mul- 
is AB and CD together of E and F to- 
r. 

erefore, if any magnitudes, how many soever, be equi- 

les of as many, each of each, whatsoever multiple any 

^ them is of its part, the same multiple shall all the first 

' udes be of all the other: * For the same demonstration 

I 2 ' holds 



it6 



THE ELEMENTS 



Book V. ( holds in anv number of maguuuaes, which was here applied 
''•^•^^ t to two.' Q. E. D. 



PROP. II. THEOR. 



I 



If the fii-st magnitude be the same multiple of th( 
second that the third is of the fourth, and the fiftl 
the same multiple of the second that the sixth is o 
-the fourth ; then shall the first together with tb 
fifth be. the same multiple of the second, that th 
third together with the sixth is of the fourth. 

Let AB the first, be the same nt^ultiple of C the second, th; 
-DE the third is of F the fourth ; and BG the fifth, the san 



3 



multiple of C the second, that EH 
the sixth is of F the fourth : Then is 
AG the first, together with the fifth, 
the same multiple of C the second, 
that'DH the third, together with 
the sixth, is of F the fourth. 

Because AB is the same multiple 
of C, that DE is of F i there are as 
many magnitudes in AB equal to C, 
as there are in DE equal to F : In like 
manner, as many as there are in BG equal to C, so many 
there in EH equal to F : As many, then, as are in the wl 
AG equal to C, so many are there in the whole DH eqy^ 
F J therefore AG is the same multiple of C, that DH isp! 
that is, AG the first and fifth together, is 
the same multiple of the second C, that 
DH the third and sixth together is of the 
fourth F. If therefore, the first be the 
same multiple, &c. Q^ E. D. 



Dl 



E 



H 



A 



Cor. * From this it is plain, that, if any 
'number of magnitudes AB, BG, GH, 

* be multiples of another C j and as many 

* DE, EK, KL be the same multiples of 

* F, each of each j the whole of the first, 
•viz. AM, is the same multiple of C, 

* that the whole of the last, viz. DL, is 
«ofF* 



B 



G 



D 



:et 



X- 



K C tl 



OF EUCLID. 



PROP. III. THEOR. 




F the first be the same multiple of the second, 
which the third is of the fourth ; and if of the first 
and third there be taken equimultiples, these shall 
be equimultiples, the one of the second, and the 
other of the fourth. 



Let A the first, be the same multiple of B the second, that 
C the third is of D the fourth ; and or A, C let the equimul- 
tiple- EF, GH be taken : Then EF is the same multiple of 
B, that GH ts of D. 

Because EF is the same multiple of A, that GH is of C, 
there arc as many magnitudes in EF equal to A, as are in GH 
equal to C : Let EF be di- 

ided into the magnitudes ■^' 
EK, KF, each equal to A, H* 

and GH into Gl,, LH, 
fleach equal to C : Tfie num- 
ber therefore of the magni- 
itudes EK, KF, s^all be 

equal to the number ot the • t- 

others GL, LH : And be- 
cause A is the same multi- 
ple of B, that C is of D, 
and that EK is equal to A, 
land GL to C ; therefore 
lEK is the same multiple E A B 
ofB,thatGLisofD: For 

I the same reason, KF is the same multiple of B, that LH is of 
;D ; and so, if there be more parts m EF, GH equal to A, C : 
I Because, therefore, the first EK is the same multiple of the 
second B, which the third GL is of the fourth D, and that 
the fifth KF is the same multiple of the second B, which the 
sixth LH is of the fourth D j EF the first together with the 
I fifth, is the same multiple* of the second J3, which GH the » 2. 5. 
' third, together with the sixth, is of the fourth D. If, there- 
f fore, the first, &c. Q. E. D. 

ft 'r 



C D 



ii8 



THE ELEMENTS 



SceX. 



UookV. 

PROP. IV. THEOR. 

XF the first of four magnitudes has the same ratio 
to the second which the third has to the fourtli ; 
then any equimultiples whatever of the first and 
third shall have the same ratio to any equimultiples 
of the second and' fourth, viz. ' the equimultiple of 
' the first shall have the same ratio to that of the 
' second, which the equimultiple of the thii-dhas to 
' that of the fourth.' 

Let A the first, have to B the second, the same ratio which 
the thrrd C has to the fourth D i and of A and C let there b« 
taken any equimultiples whatever 
E, F ; and of B and D any equi- 
multiples whatever G, H : Then 
E has the same ratio to G, which 
F has to H. 

Take of E and F any equimul- 
tiples whatever K, L, and of G, 
H, any equimultiples whatever M, 
N : Then, because E is the same 
multiple of A, that F is of C ; 
and of E and F have been taken 
equimultiples K, L ; therefore K 

is the same multiple of A, thatL -rr ip /» »> r' 
is of C*: For the same reason, M -"^ Xi A £ G 
is the s^me multiple of B, that N 
js of D : And because, as A is to L f C D IX K 

» Hypoth. B, so is C to D\ and of A and 
C have been taken certain equi- 
multiples K,.L : and of B and D 
have been taken certain equimul- 
tiples M, N: if therefore K be 
greater than M, L is greater than 
N : and if equal, equal ; if less, 

' 5. def. 5. less<=. And K, L, are any equi- 
multiples whatever of E, F j and 
M, N any whatever of G, H : As 
therefore E is to G, so is*^ F to H. 
Therefore, if the first, &c. Q^ E, D. 

Cor. Likewise, if the first has the same ratio to the second, 
which the third has to the fourth, then also any equimultiples 

whatever 



J. j. 



^ 



.1 



£«« N. 



i 



OF EUCLID. 

whatever of the first and third have the same ratio to the se- Bo 
cond and fourth : And in like manner, the first and the third 
have the same ratio to any equimultiples whatever of the se- 
cond and fourth. 

Let A the first, have to B the second, the same ratio which 
the third C has to the fourth D, and of A and C let E and F 
be any equimultiples whatever ; then E is to B, as F to D. 

Take of E, F any equimultiples whatever K, L, and of B, 
Dany equimultiples whatever G, H ; then it may be demon- 
strated, as before, that K is the same multiple of A, that L is 
of C : And because A is to B, as C is to D, and of A and C 
certain equimultiples have been taken, viz. K and L ; and of 
B and D certain equimultiples G, H ; therefore, if K be greater 
than G, L is greater than H ; and if equal, equal ; if less, less*^ : 
And, K, L are any equimultiples of E, F, and G, H any what- 
ever of B, D; as therefore E is to B, so is F to D; And in 
the same way the other ca§e is demonstrated. 




def. 5. 



PROP. V. THEOR. 



J.F one magnitude be the same multiple of another, s« n. 
which a magnitude taken from the first is of a mag;- 
nitude taken from the other; the remainder shall 
be the same multiple of the remainder, that the 
whole is of the whole. 

Let the magnitude AB be the same multiple 

of CD, that AE taken from the first, is of CP 

taken from the other j the remainder EB shall 

be the same multiple of the remainder FD, that 

;the whole AB is of the whole CD. 

Take AG the same multiple of FD, that 
AE is of CF : therefore AE is* the same mul- 
tiple -of CF, that EG is of CD: But AE, by 
the hypothesis, is the same multiple of CF, that 
AB is of CD : Therefore EG is the same mul- 
tiple of CD that AB is of CD : wherefore EG 
is equal to AB''. Take from them the common 
magnitude AE;the remainder AG is equal to 
-the remainder EB. Wherefore, since AE is 
the same multiple of CF, that AG isofFD, 
and that AG is equal to EB ; therefore AE is the same multiple 
ofCF, that EB is of FD: But AE is the same multiple of CF, 

14 that 



A- 



E 



B 



M, AX.5J. 



120 

Book V. 



SceN. 



i Ax. 5, 



A 



K 



THE ELEMENTS 

that A B is of CD ; therefore EB is the same multiple of FD, 
' that A B is of CD. Therefore, if any magnitude, &c. Q. E. D. 

PROP. VI. THEOR. 

XF two magnitudes be equimultiples of two others, 
and if equimultiples of these be taken from the first 
two, the remainders arc either equal to these others, 
or equimultiples of them. 

Let the two magnitudes AB, CD be equimultiples of the 
two E, F, and AG, CH taken from the first two be equimul- 
tiples of the same E, F ; the remainders GB, HD are either 
equal to E, F, or equimultiples of them. 

First, let GB be equal to E ; HD is 
' equal to F : Make CK equal to F ; andbe- 
bause AG is the sam^e multiple of E, that 
CH is of F, and that GB is equal to E, 
and CK to F ; therefore AB is the same 
multiple of E, that KH is of F. But AB, 
by the hypothesis, is the same multiple of 
E that CD is of F ; therefore KH is the 
same multiple of F, that CD is of F ; 
wherefore KH is equal to CD* : Take 
away the common magnitude CH, then 
the remainder KC is equal to the remain- 
der HD : But KC is equal to F; HD therefore is equal to F. 

But let GB be a multiple of E ; then 
HD is the same multiple of F : Make 
CK the same multiple of F, that GB is 
of E: And because AG is the same mul- 
tiple of E, that CH is of F; and GB the 
same multiple of E, that CK is of F j 
therefore AB is the same multiple of E, 
that KH is of F'': But AB is the same 
multiple of E, that CD is of F j therefore 
KH is the same multiple of P', that CD is 
of it ; Wherefore KH is equal to CD* : 
Take away CH from both ; therefore the 
remainder KC is equal to the remainder 
HD : And because GB is the same multiple of E, that KC is 
ofF, and that KC is equal to HD ; therefore HD is the same 
multiple of F, that GB is of E. If therefore two magnitudes, 
kc. Q. E. D. 



C 



H 



E D E F 



X. 



A 



H 



E D E F 



OF EUCLID. 



PROP. A. THEOR. 




IF the first of four magnitudes has to the second, 
the same ratio which the third has to tiie fourth ; 
then, if the first be greater than the second, the 
third is also greater than the fourth ; and if equal, 
equal ; if less, less. 

Take any equimultiples of each of them, as the doubles of 
each J then, by def. 5th of this book, if the double of the first be 
greater than the double of the second, the double of third is 
greater than the double of the fourth : but, if the first be greater 
than the second, the double of the first is greater than the double 
of the second ; wherefore also the double of the third is greater 
than the double of the fourth ; therefore the third is greater than 
the fourth: In like manner, if the first be equal to the second, 
or less than it, the third can be proved to be equal to the fourth, 
or less than it. Therefore, if the first, &c. Q^ E. D. 



SeeN. 



PROP. B. THEOR. 

\ F four magnitudes are proportionals, they are pro- sceN. 
portionals also when taken inversely. 

If the magnitude A be to B, as C is to D, then also inverse- 
ly B is to A, as D to C. 

Take of B and D any equimultiples 
whatever E and F ; and of A and C any 
equimultiples whatever G and H. First let 
£ be greater than G, then G is less than E ; 
and because A is to B, as C is to D, and 
•f A and C, the first and third, G and H 
are equimultiples; and of B and D, the se- 
cond and fourth, E and F are equimulti- 
ples; and that G is less than E, H is also GAB 
*less than F ; that is, F is greater than H ; tt Q Y) 
if therefore E be greater than G, F is greater 
than H: In like manner, if E be equal to 
G, F may be shown to be equal (Q H ; and 
if less, less ; and E, F, are any equimul- 
tiples whatever of B and D, and G, H i . 
any whatever of A and C ; therefore, as B 

is 



•5.4ef.5. 



122 THE ELEMENTS 

£oo;<v. is to A, SO IS D to C. If, then, four magnitudes, &c. 
^-^^Q,E. D. 

PROP. C. THEOR. 

s«eN. J F tlie first be the same multiple of the second, or 
the same part of it, that the third is of the fourth; 
the first is to the second, as the third is to the fourth. 

Let the first A be the same multiple of B 
the second, that C the third is of the fourth 
D : A is to B as C is to D. 

Take of A and C any equimultiples what- 
ever E and F j and of B and D any equi- 
multiples whatever G and H : Then, because 
A is the same multiple of B that C is of D j 
and that E is the same multiple of A, that 
F is 'of C : E is the same multiple of B, that 

•3.5. F is of D*i therefore E and F are the same 
multiples of B and D : But G and H are equi- 
multiples of B and D: therefore, if E be a 
greater multiple of B than G is, F is a greater 
multiple of D than H is of D ; that is, if E 
be greater than G, F is greater than H : 
In like manner, if E be equal to G, or less, 
F is equal to H, or less than it. But E, F 
are equimultiples, any whatever, of A, C, 
and G, H, any equimultiples whatever of B, 

• 5. dcf. 5. D. Therefore A is to B, as C is to D^ 

Next, Let the first A be the same part 
of the second B,^ that the third C is of 
the fourth D : A is to B, as C is to D : 
For B is the same multiple of A, that D 
is of C: wherefore by the preceding 
case, B is to A, as D is to C ; and in- 
versely"^ A is to B as C is to D : There- 
fore, if the first be the same multiple, &c. 
Q. E. D. 



A B C D 
E G FH 



•B.5, 



A B C D 



OF EUCLID. 



PROP. D. THEOR. 




If the first be to the second as the third to the^^'* 
fourth, and if the first be a multiple or part of the 
second j the third is the same multiple, or the same 
part of the fourth. 

Let A be to B, as C is to D; and first let A be a multiple 
of B ; C is the same multiple of D. 

Take E equal to A, and whatever mul- 
tiple A or E is of B, make F the same mul- 
tiple of D: Then, because A is to B, as C 
is to D ; and of B the second, and D the 
fourth equimultiples have been taken E and 
F ; A is to E, as C to F - : But A is equal 
to E, therefore C is equal to F ^ : and F 
is the same miJtiple of D, that A is of B. A B C D 
Wherefore C is the same multiple of D, 
that A is of B. 

Next, Let the first A be a par£of the se- 
cond B ; C the third is the same part of the 
fourth D. 

Because A is to B, as C is to D ; then, 
inversely, B is= to A, as D to C: But A is 
a part of B, therefore B is a multiple of A; 
and, by the preceding case, D is the same 
multiple of C ; that is, C is the same part of D, that A is of 
B : Therefore, if the first, &c. Q; E. D. 



» Cor. 4. 5. 
"A. 5. 



See the fi- 
gure at th« 
foot of tb« 
preceding 
page- 



PROP. Vn. THEOR. 

JlLQUAL magnitudes have the same ratio to the 
same magnitude ; and the same has the same ratio 
to equal magnitudes. 

Let A and B be equal magnitudes, and C any-other. A and 
B have each of them the same ratio to C, and C has the same 
ratio to each of the magnitudes A and B. 

Take of A and B any equimultiples whatever D and E,and 

of 



124 



THE ELEMENTS 



SceN. 



E 



) A 

B 



C F 



Book V. of C any multiple whatever F : Then, because D is the same 

'^'^'^'^^'^ multiple of A, that E is of B, and that A is 

*l. Ax. 5. equal to B ; D is* equal to E : Therefore, if 

^ D be greater than F, E is greater than F ; and 

if equal, equal ; if less, less : And D, E are 

any equimultiples of A, B, and F is any mul- 

*5 def. 5. tiple of C. Therefore^, as A is to C, so is 
B to C. 

Likewise C has the same ratio to A, that it 
has to B : For, having made the same con- 
stru£lion, D may in like manner be shown 
equal to E: Therefore, if F be greater than 
D, it is likewise greater than E ; and ifequal, 
equal ; if less, less : And F is any multiple 
whatever of C, and D, E are any equimul- 
tiples whatever of A, B. Therefore C is to 
A, as C is to B^ Therefore, equal magni- 
tudes, &c. Q. E. D. 

PROP. Vm. THEOR. 

vJF unequal magnitudes, the greater has a greater 
ratio to the same than the less has ; and the same 
magnitude has .a greater ratio to the less, than it 
has to the greater. 

Let AB, B'C be unequal magnitudes, of which AB is the 
greater, and let D be any magnitude 
whatever: AB has a greater ratio to D 
than BC to D : And D has a greater 
ratio to BC than unto AB. 

If the magnitude which is not the 
greater of the tv/o AC, CB, be not less 
than D, take EF, FG, the doubles pf 
AC, CB, as in Fig. i. But if that which 
is not the greater of the two AC, CB 
be less than D (as in Fig 2 and 3.) this 
magnitude can be multiplied, so as to 
become greater than D, whether it be 
AC, or CB. Let it be multiplied, until 
it become greater than D, and let the 
other be multiplied as often; and let EF 
be the multiple thus taken of AC, and 
FG the same multiple of CB : Therefore 
EF and FG are each of them greater than 

D: And 



E 



Fig. 



A 



I. 



G 

L 



K. H D 



OF EUCLID. 



D : And in every one of the cases, take H the double of D, K 
its triple, and so on, till the multiple of D be that which first 
becomes greater than FG : Let L be that multiple of D which 
is first greater than FG, and K the multiple of D which is 
next less than L. 

Then, because L is the multiple of P, which is the first that 
becomes greater than FG, the next preceding multiple K is 
not greater than EG j that is, FG is not less than K : And since 
EF is the same multiple of AC, that FG is of CB j FG is the 
same multiple of CB ; that EG is of AB* ; wherefore EG and 
FG are equimultiples of AB and CB : And it was shown, that 
FG was not less than K, 

and, by the construe- Fig. 2. Fig. 3. 

tion,EF, is greater than Y.\ E 

D ; therefore the whole 
EG is greaterthanKand F 
D together : But K, to- 
gether with D, is equal 
to L ; therefore EG is 
greater than L; but FG 
is not greater than L ; 
and EG, FG are equi- 
multiples of AB, BC, Q Y^ C 
and L is a multiple of 
D; therefore b AB has L K H D 
to D a greater ratio 
than BC has to D. L K D 

Also D has to BC a 
greater ratio than it lus 
to AB : For, having 
made the same con- 
stru£bion, it may be 
shown, in like manner, 
that L is greater than 
FG, but that it is not greater than EG : and L is a multiple 
of D ; and FG, EG are equimultiples of CB, AB ; therefore 
D has to CB a greater ratio'' than it has to AB. Wherefore, 
•f unequal magnitudes, &c. Q^ E. D. 



A 
C 



1.5. 



det 




SwN. 



THE ELEMENTS 

PROP. IX. THEOR. 

IVaagnitudes which have the same ratio to the 
same iliagnitude are equal to one another; and 
those to which the same magnitude has the same 
ratiQ are equal to one another. 

Let A, B have each of them the same ratio to C : A is 
equal to B : for, if they are not equal, one of them is greater 
than the other ; let A be the greater ; then, by what was shown 
in the preceding proposition, there are some equimultiples of 
A and B, and some multiple of C such, that the multiple of A 
is greater than the multiple of C, but the multiple of B is not 
greater than that of C. Let such multiples be taken, and let 
I), E, be the equimultiples of A, B, and F the multiple of C, 
so that D may be greater than F, and E not greater than F : 
But, beeause A is to C, as B is to C, and 
of A, B, are taken equimultiples D, E, 
and of C is taken a multiple F j and that Vv 

D is greater than F ; E shall also be greater 
- 5. def. 5. than F^ ; but E is not greater than P" ; ^ 
which is impossible J A therefore and B 
are not unequal ; that is, they are equal. 

Next, let C have the same ratio to each 
of the magnitudes A and B ; A is equal 
to B : For, if they are not, one of them is ■.> 
.greater than the other ; let A be the great- 
er ; therefore, as was shown in Prop. 8th, 
there is some multiple F of C, and some 
equimultiples E and D, of B and A such, 
that F is greater than E, and not greater than D ; but because 
C is to B, as C is to A, and that F, the multiple of the first, 
is greater than E, the multiple of the second ; F the multiple 
of the third, is greater than D, the multiple of the fourth' : 
But F is not greater than D, which is impossible. Tliere- 
fore A is equal to B. Wherefore, magnitudes which, &c. 
Q,E.D. 



OF EUCLID. 

PROP. X. THEOR. 




X HAT magnitude which lias a greater ratio than s«ex 
another has unto the same magnitude, is the s:reat- 
er of the two : And that magnitude to which the 
same has a greater ratio than it has unto another 
magnitude is the lesser of the two. 

Let A have to C a greater ratio than B has to C ; A is great- 
er than B : For, because A has a greater ratio to C, than B has 
to C, there are* some equimultiples of A and B, and some *7. def. 5. 
multiple of C such, that the multiple of A is greater than the 
multiple of C, but the multiple of B is not greater than it : Let 
them be taken, and let D, E be equir ul- 
tiples of A, B, and F a multiple of C such, 
that D is greater than F, but E is not J) 

greater than F : Therefore D is greater 
than E : And, because D and E are equi- 
multiples of A and B, and D is greater A 
than E ; therefore A is^ greater than B. 

Next, let C have a greater ratio to B ^1 V **.Ax.5. 

than it has to A ; B is less than A : For* 
there is some multiple F of C, and some 
equimultiples E and D of B and A such, B 
that F is greater than E, but is not greater 
than D : E therefore is less than D ; and 
because E and D are equimultiples of B 
and A, therefore B is'' less than A. That magnitude, there- 
fore, &c. Q. E. D. 



PROP. XL THEOR. 

Jtv ATIOS that are the same to the same ratio, are 
the same to one another. 

Let A be to B as C is to D; and as C to D, so let E be to 
F J A is to B, as E to F. 

Take of A, C, E, any equimultiples whatever G, H, Kj and 
of B, D, F, any equimultiples whatever L,M, N. Therefore, 
since A is to B, as C to D, and G, H are taken equmultiples of 

A,C, 



128 ■ THE ELEMENTS 

OOK V 



» :). def. 5. 



$ , C, and L, M, of B,D ; if G be greater than L, H is greater 
than Mj and if equal, equal; and if less, less\ Again, be- 
cause C is to D, as E is to F, andH,Kare taken equimultiples 
of C, E ; and M, N, of D, F ; if H be greater than M, K is 
greater than N ; and if equal, equal ; and if less, less : But if 

G H K 



A C E ■ 

B I> F 

L M N 

G be greater than L, it has been shown that H is greater than 
M, and if equal, equal ; and if less, less ; therefore, if G be 
greater than L, K is greater than N ; and if equal, equal ; and 
if less, less : And G, K are any equimultiples whatever of A, 
E ; and L, N any whatever of B, F : Therefore, as A is to B, 
so is E to F\ Wherefore, ratios that, Sec. Q. E. D. 



PROP. XII. THEOR. 

JlF any number of magnitudes be proportionals, as 
one of the antecedents is to its consequent, so shall 
all the antecedents taken together be to all the con- 
sequents. 

Let any number of magnitudes A, B, C, D, E, F, be pro- 
portionals ; that is, as A is to B, so C to D, and E to F : As A is 
to B, so shall A, C, E together be to B, D, F together. 

Take of A, C, E any equimultiples whatever G, H, K ; 

G H K 

A C F 



B D— , F 

L M-^^ N- 



and of B, D, F any equimultiples whatever L, M, N : Then, 
because A is to B, as C is to Djand as E to F ; and that G, H, 

K 



OF EUCLID. 129 

K are equimultiples of A, C, E, and L, M, N, equimultiples of BookV. 
B, D, F ; if G be greater than L, H is greater than M, and K iS.Def. 5. 
greater than N ; and if equal, equal ; and if less, less*. Where- 
fore, if G be greater than L, then G, H, K together are greater 
than L, M, N together j and \i equal, equal ; and if less, less. 
And G, and G, H, K, together are any equimultiples of A, and 

A, C, E togetner ; because if there oe any number of magni- 
tudes equimultiples of as many, each of each, whatever mul- 
tiple of one of them is of its part, the same multiple is the % 
whole of the whole": For the same reason L, and L, M, N ^" ^' 
are any equimultiples of B, and B, D, F : As therefore A is to 

B, so are A, C, E, tosether to B, D, F together. Where^ 
fore, if any number, &c. Q. E. D. 

PROP. XIII. THEOR. 

JLF the first has to the second the same ratio which ^^'' 
the third has to the fourth, but the third to the 
fourth a greater ratio than the fifth has to the sixth; 
the first shall also have to the second a greater ra- 
tio than the fifth has to the sixth. 

Let A the first, have the same ratio to B the second, which 
C the third, has to D the fourth, but C the third, to D the 
fourth, a greater ratio than E the fifth, to F the sixth : Also 
the first A shall have to the second B a greater ratio than the 
fifth E to the sixth F. 

Because C has a greater ratio t« D, than E to F, there are 
some equimultiples of C and E, and some of D and F such, 
that the multiple of C is greater than the multiple of D, but 

M G H 



B D F 

N K L 

the multiple of E is not greater than the multiple of F» : Let • T. drf. 5. 
Isuch be taken, and of C, E let G, H be equimultiples, and K, 
|iL equimultiples of D, F, so that G be greater than K, but H 
aot greater than L ; and whatever multiple G is of G, take M 
:he same multiple of A ; and whatever multiple K isof D, take 
M the same multiple of B : Then, because A is to B, as C to 

K D, and 



I 

1 



130 



THE ELEMENTS 



'5.Def,5. 



Def. 5. 



Book V. J)^ jfjd of A and C, M and G are equimultiples ; And of B and 
D, N and K are equimultiples ; if M be greater than N, G is 
greater than K ; and if equal, equal ; and if less, less'' ; but G 
is greater than K, therefore M is greater than N : But H is not 
greater than L ; and M, H are equimultiples of A, E ; and N, 
L equimultiples of B, F : Therefore A has a greater ratio to B, 
than E has to F^ Wherefore, if the first, &c. Q^ E. D. 

Cor. And if the first have a greater ratio to the second, than 
the third has to the fourth, but the third the same ratio to the 
fourth, which the fifth has to the sixth j it may be demon- 
strated, in like manner, that the first has a greater ratio to the 
second, than the fifth has to the sixth. 

PROP. XIV. THEOR. 

See N. J[f the first has to the second, the same ratio which 
the third has to the fourth ; then, if the first be 
greater than the third, the second shall be greater than 
the fourth ; and if equal, equal ; and if less, less. 

Let the first A have to the second B, the same ratio which 
the third C, has to the fourth D j if A be greater than C, B is 
greater than D. 

Because A is greater than C, and B is any other magnitude^ 
« 1. 5. A has to B a greater ratio than C to B'' ; But, as A is to B, sq 



3 



A B C D A B C D A B C D 

is C to D i therefore also C has to D a greater ratio than C 
has to B^ But of two magnitudes, that to v^ich the same 
has the greater ratio is the lesser^ Wherefore D is less than 
B ; that is, B is greater than D. 

Secondly, if A be equal to C, B is equal to D : For A is to 
B, as C, that is A, to D : B therefore is equal to D*". 

Thirdly, if A be less than C, B shall be less than D : For C 
is greater than A, and because C is to D, as A is to B, D is 
greater than B, by the first case ; wherefore B is less than Di 

Therefore, if the first, &c. Q.. E. D. 



» 13. 5. 
« 10. 5. 

*9.5. 



OF EUCLID. 




PROP. XV. THEOR. 



IVIagnitudes have the same ratio to one ano- 
ther which their equimultiples have- 
Let AB be the same multiple of C, that DE is of F j C is f 
F, as AB to DE. 

Because AB is the same multiple of C, thatDE is of F; there 
areas many magnitudes in AB equal to C, 
as there are in DE equal to F : Let AB be 
divided into magnitudes, each equal to C, tx 

viz. AG, GH, HB ; and DE into magni- 
tudes, each equal to F, viz. DK, KL, LE : 
Then the number of the first AG, GH, H B, K 

shall be equal to the number of the last DK, 
KL, LE : And because AG, GH, HB are 
all equal, and that DK, KL, LE, are also 
equal to one another ; therefore AG iS to 
DK, as GH to KL, and as HB to LE M B C E F 
And as one of the antecedents to its conse- 
quent, so are all the antecedents together to all the consequents 
together'' ; wherefore, as AG is to DK, so is AB to DE : But " ^2- 5. 
AG is equal to C, and DK to F : Therefore, as C is to F, 
so is AB to DE. Therefore magnitudes, &c. Q^ E. D. 



H- 



PROP. XVL THEOR. 



xFfour magnitudes of the same kind be proportion- 
als, they shall also be proportionals when taken al- 
ternately. 

Let the four magnitudes A, B, C, D, be proportionals, viz. 
as A to B, so C to D : They shsill also be proportionals when 
taken alternately ; that is, A is to C, as B to D. 
, Take of A and B any equimultiples whatever E and F ; and 
•f C and D take any equimultiples whatever G and H : and 

K 2 because 




THE ELEMENTS 

_^^^ because E is the same multiple of A, that F is of B, and that 

M3.5. magnitudes have the same ratio to one another which their 
equimultiples have^j therefore A is to B, as E is to F : But as 
A is to B, so is C to 

»il. 5. D : Wherefore as C E" — G 

is to D, so *> is E to p 

F: Again, because ^ ^ 

G, H are equimul- i> T) 

tiples of C, D, as C 

is to D, so is G to F H 

H» ; but as C is to 

D, so is E to F. Wherefore, as E is to F, so is G to H^ But 
when four magnitudes are proportionals, if the first be greater 
than the third, the second shall be greater than the fourth ; and 

' U. 5. If equal, equal j if less, less"^. Whtrefore, if E be greater than 

G, F likewise is greater than H : and if equal,, equal ; if less, 

less : And E, F arc any equimultiples whatever of A, B ; and 

^ G, H any whatever of C, D. Therefore A is to C as B to 

*5. Def. 5. Dd. If then four magnitudes, &c. Q, E. D. 



SccN. 



PROP. XVII. THEOR. 

JLF magnitudes, taken jointly,' be proportionals, 
they shall also be proportionals when taken sepa- 
rately; that is, if two magnitudes together have to 
one of them the same ratio which two others have 
to one of these, the remaining one of the first two 
shall have to the other the same ratio which the 
remaining one of the last two has to the other of 
these. 

Let AB, BE, CD, DF be the magnitudes taken jointly 
which are proportionals'; that is, as AB to BE, so is CD to 
DF ; they shall also be proportionals taken separately, viz. 
as AE to EB, so CF to FD. 

Take of AE, EB, CF, FD any equimultiples whatever GH, 
HK, LM,MN; and again, of EB, FD take any equimultiples 
whatever KX, NP : And because GH is the same multiple of 
AE, that HK is of E B, wherefore GH is the same multiple* of 
AE, that GK is of AB : But GH is the same multiple of AE, 
that LMi^ ofCFj wherefore GK is the same multiple of AB, 

I that 



OF EUCLID. 



^33 



X 



K 



*2. 5. 



that LVI is of CF. Again, because LM is the same mukrple of Book v 
CF, that MN is of FD; therefore LM is the same multiple^'^'^^ 
of CF, than LN is of CD : But LM was shown to be the same 
multiple of CF, that GK is of AB ; GK therefore is the same 
multiple of A H, that LN is of CD ; that is, GK, LN areequi- 
multiples of AB, CD. Next, because HK is the same multiple 
ofEB, thatMNisofFD; and that KX is 
also the same multiple of EB, that NP is 
of FD; therefore HX is the same multiple 
*'of EB, that MP is of FD. And because 
AB is to BE, as CD is to DF, and that of 
AB and CD, GK. and LN are equimul- 
tiples, and of EB and FD, HX and MP 
are equimultiples ; if GK be greater than 
HX, then LN is greater than MP ; and 
if equal, equal ; and if less, Iess<^ ; But if H 
GH be greater than KX, by adding the 
common part HK to both, GK is greater 
than HX ; wherefore also LN is greater 
than MP ; and by talcing away MN from 
both, LM is greater than NP : There- 
fore, if GH be greater than KX, LM is 
greater than NP. In like manner it may 
be demonstrated, that if GH be equal to KX, LM likewise is 
equal to NP; and if less, less : And GH, LM are any equi- 
multiples whatever of AE, CF, and KX, NP are any what- 
ever of EB, FD. Therefore^ as AE is to EB, so is CF to 
¥D. If then magnitudes, &c. Q^ E. D. 



B 

E 



D 
F 



N 



M' 



5. De£ 5. 



G A C L 



PROP. XVIII. THEOR. 

J-F magnitudes, taken separately, be proportionals, 
they shall also be proportionals "when taken jointly, 
that is, if the first be to the second, as the third to 
the fourth, the first and second together shall be to 
the second, as the third and fourth together to the 
fourth. 

LetAE, EB, CF, FD be proportionals ; that is, as AE 
to EB, so is CF to FD ; they shall also be proportionals whcQ- 
taken jointly ; that Is, as AB to BE, so CD to DF. 

Take of AB, BE, CD, DF any equimultiples whatever 
GH, HK, LM, MN : and again, of BE, DF, take any what- 
ever equimultiples KG, NP: And because KO, NP are 

K 3 equimultiples. 



Seeh. 



13+ 



THE ELEMENTS 







K 



«S.5. 



M 
P- 

N 






^°°^- equimultiples of BE, DF ; and thatKH, NMare equimultiples 

^"^""^^^ likewise of BE, DF, if KO, the multiplcof BE, be greater than 
KH, which is a multipleof the same BE, NP, likewise the mul- 
tiple of DF, shall be greater than MN, 
the multiple of the same DF ; and if IT 
KO be equal to KH, MP shall be 
cq«al to NM ; and if less, less. 

First, Let KO not be greater than 
KH, therefore NP is not greater than 
NM : And because GH, HK, are equi- 
multiples of AB, BE, and that AB is 
greater than BE, therefore GH is 

« 5. Ax. 5. greater* than HK j but KO is not 
greater than KH, wherefore GH is 
greater than KO. In like manner it B 

may be shown, that LM is greater than 
NP.. Therefore, if KO be not greater E 

than KH, then GH, the multiple of 
AB, is always greater than KO, the ^ 
multiple of BE ; and likewise LM, the 
multiple of CD, greater than NP, the multiple of DF. 

Next, Let KO be greater than KH : therefore, as has been 
shown, NP is greater than NM : And because the whole jH 
is the same multiple of the whole AB, that HK is of BE, the 
remainder GK is the same multiple of 

*5.5. the remainder AE that GH is of AB^: q 
which is the samethat LM is of CD. 
In like manner, because LM is the K 
same multiple of CD, that MN is of 
DF, the remainder LN is the same 
multiple of the remainder CF, that 
the whole LM is of the whole CD'' : K 
But it was shown that LM is the same -n 

" multiple of CD, that GK is of AE j D" 

therefore GK is the same multiple of E 

AE, thatLN isof CF; that is, GI^, 
LN are equimultiples of AE, CF : 
And because KO, NP are equimul- G A 
tiples of BE, DF, if from KO, NP, 

there be taken KH, NM, which are likewise equimultiples 
of BE, DF, the remainders HO, MP are either equal to B?2, 
DF, or equimultiples of them ^ First, let HO, MP, be 
equal to BE, DF j and because AE is to EB, as CF to FD, and 

that 



P 

M 

N 



OF EUCLID. 

that GK, LN are equimultiples of AE, CF ; GK shall be to 
EB, as LN to FD' : But HO is equal to EB, and MP to 
FD ; wherefore GK is to HO, as LN to MP. If therefore 
GK be greater than HO, LN is greater than M P ; and if 
equal, equal ; and if less^ less. 

But let HO, MP be equimultiples of EB, FD ; and because 
AE is to EB, as CF to FD, and that of AE, CF are taken 
equimultiples GK LN ; andofEB,FD,theequimultiplesHO, 
MP ; if GK be greater than HO, LN 
is greater than MP ; and if equal 
equal ; and if less, less*^ ; which was 
likewise shown in the preceding case. 
If thereforeGH be greater than KO, 
taking KH from both, GK is greater 
than HO ; wherefore also LN is 
greater than MP ; and consequently, 
adding NM to both, LM is greater 
than NP : Therefore, if GH be 
greater than KO, LM is greater than 
NP. In like manner it maybe shown, 
that if GH be equal to KO, LM is 
equal to NP ; and if less, less. And 
in the case in which KO is not greater 
than KH, it has been shown that GH 
is always greater than KO, and likewise LM than NP : But 
GH, LM are any equimultiples of AB, CD, and KO, NP 
are any whatever of BE, DF ; therefore*^, a«; AB is to BE, so 
is CD to DF. If then magnitudes, &c. Q^ E. D. 



Book V. 
^ Cor, 45. 

= A. 5. 







H 



G 



B 


D 
F 


A 


C 



M 

N 



f=; 



5. D«f. 5. 



PROP. XIX. THEOR. 



JLF a whole magnitude be to a whole, as a magui- See n. 
tude taken from the first, is to a magnitude taken 
from the other; the remainder shall be to the re- 
mainder, as the whole to the whole. 

Let the whole AB, be to the whole CD, as AE, a magni- 
tude taken from AB, to CF, a magnitude taken from CDj 
the remainder EB shall be to the remainder FD, as the whole 
AB to the whole CD. 

Because AB is to CD, as AE to CF : likewise, alternately*, » le 5, 
K4 BA - 



'36 



Book V. 



" 17. 5. 



THE ELEMENTS 

BA is to AE, as DC Is te CF : and because, if 
magnitudes, taken jointly, be proportionals, they 
are also proportionals^ when taken separately j 
therefore, as BE is to EA, so is DF to FC, and 
alternately, as BE is to DF, so is EA to FC : 
But, as AE toCF, so by the hypothesis, is AB to 
CD J therefore also BE, the remainder, shall be 
to the remainder DF, as the whole AB to the whole 
CD : Wherefore, if the whole, &c. Q. E. D. 

Cor. If the \V^hole be to the whole, as a mag- 
nitude taken from the first,is to a magnitude taken B D 
from the other ; the remainder likewise is to the 
remainder ; as the magnitude taken from the first to that taken 
from the other ; The demonstration is contained in the pre- 
ceding. 

PROP. E. THEOR. 

1 F four magnitudes be proportionals, they are also 
proportionals by conversion ; that is, the first is to 
its excess above the second, as the third to its ex* 
cess above the fourth. 



» 17. 5. 
•>B. 5. 
« IS. 5. 



Let AB be to.BE, as CD to DF j then BA 
is to AE, as DC to CF. 

Because AB is to BE, as CD to DF, by divi- 
sion*, AE is to EB, as CF to FD ; and by in- 
version'', BE is to EA, as DF to FC. Where- 
fore, by composition", BA is to AE, as DC is 
to CF : If, therefore, lour, &c. Q^ E. D. 



E 



B n 



SceN. 



PROP. XX. THEOR. 

If there be three magnitudes, and other three, 
which, taken two and two, have the same ratio; if" 
the first be greater than the third, the fourth shall 
be greater than the sixth ; and if equal, equal ; and 
if less, less. 

Let 



OF EUCLID. 



^37 



D E F 



Let A, B, C be three magnitudes, and D, E, F other three, BookV^ 
which, taken two and two, have the same ratio, viz. as A is to ^■^""'''^^ 
B, so is D to E ; and as B to C, so is E to 
F. If A be greater than C, D shall be greater 
than F : and, if equal, equal ; and if less, 
less. 

Because A is greater than C, and B is any 
other magnitude, and that the greater has to 
the same magnitude a greater ratio than the less 
has to it^ ; therefore A has to B a greater ratio 
than C has to B : Bui as D is to E, so is A to A B 
B ; therefore'' D has to E a greater ratio than 
C to B ; and because B is to C, as E to F, by 
inversion, C is to B, as F is to E : and D 
was shewn to have to E a greater ratio than C 
to B ; therefore D has to E a greater ratio 
than F to E^ But the magnitude which has 
a greater ratio than another to the same magnitude, is the 
greater of the two** : D is therefore greater than F. 

Secondly, Let A be equal to C ; D shall be equal to F 
cause A and C are equal to one 
another, A is to B, as C is to B^ : 
But A is to B, as D to E ; and C is 
to B, as F to E J wherefore D is to 
E, as F to Ef ; and therefore D is 
equal to Fs. 

Next, Let A be less than C ; D 
shallbe less than F: For C is greater ^ 
than A, and, as was shown in the 
first case, C is to B, as F to E, 
and in like manner B is to A, as E 
to D ; therefore F is greater than 
D, by the first case ; aad therefore 
D is less than F. Therefore, if there be three, &c. Q. E. D. 



Be- 



A B C 



Jb 



A 
D 



B C 

E F 



8.5. 



* 13.5. 



"=Cor. 13.5. 



10. 5. 



7,5. 



ni.5. 

'9.5. 



PROP. XXL THEOR. 

IF there be three magnitudes, and other three, SceN. 
whicli have the same ratio taken two and two, but 
in a cross order; if the first magnitude be greater 
than the third, the fourth shall be greater than the 



sixth ; and if equal, equal ; and if less, less. 



Let 



^3? 



THE ELEMENTS 



•8. 5. 
» 13. 5 



'Cor. 13. 5 



« 10.5 

« 7. 5. 

fll. 5 
e 9.5. 



BookV. Let A, B, C be three magnitudes, and D, E, F other three, 
^'^''^'^^ which have the same ratio, taken two and two, but in a cross 
order, viz. as A is to B, so is E to F, and as B 
is to C, so is D to E. If A be greater than C, 
D shall be greater than F ; and if equal, equal j 
and if less, less. 

Because A is greater than C, and B is ariy 
other magnitude, A has to B a greater ratio* 
than C has to B : But as E to F, so is A to B : 
therefore^ E has to F a greater ratio than C to x 
B : And because B is to C. as D to E, by inver- ^ IJ C 
sion, C is to B, as E to D : And E was' shown 
to have to F a greater ratio than C to B ; there- - J 
fore E has to F a greater ratio than E to 0*= ; 
but the magnitude to wliich the same has a 
greater ratio than it has to another, is the lesser 
of the two** : F therefore is less than D j that 
is, D is greater than F. 

Secondly, Let A be equal to C ; D shall be equal to F. Be- 
cause A and C are equal, A is^ to B, as C is to B : But A is 
to B, as E to F ; and G is to B, 
as E to D ; wherefore E is to F 
as E to D^ ; and therefore D is 
equal to F^. 

■ Next, Let A be" less than C i 
D shall be less than F : For C 
is greater than A, and, as was 
shown, C is to B, as E to D, 
and in like manner B is to A, D 
as F to E ; therefore F is greater 
than D, by case first; and there- 
fore, D is less than F. There- 
fore, if there be three, &c. 
Q, E. D. - 



A B C. A JB C 



D E F 



' PROP. XXn. THEOR. 

SteN. If there be any number of magnitudes, and as many 

others, which, taken two and two in order, have 

the same ratio ; the first shall liave to the last of the 

first magnitudes the same ratio which tlie first of the 

others has to the last. N. B. This is usually cited 

hi/ the xvords " e.v ccquali,'" oi', "e.i" cequo,'' 

First, 



OF EUCLID. 



^39 



D E F 
H L N 



»4.5. 



First, Let there be three magnitudes A, B, C, and as many ^°^^" 
others D, E, F, which, taken two and two, have the same ^■^'v^' 
ratio i that is, such that A is to B as D to E j and as B is to 

C, so is E to F ; A shall be to C, as D to F. 

Take of A and D any equimultiples whatever G and H ; 
and of B and E any equimultiples 
whatever K and L ; and of C and' 
F any whatever M and N : Then, 
because A is to B, as D to E, and 
that G, H are equimultiples of A, 

D, and K, L equimultiples of B, A "R r* 
E ; as G is to K, so is* H to L : "^ 
For the same reason, K is to M, n rr vr 
as L to N ; and because there are T" 
three magnitudes G, K, M, and 
other three H, L, N, which, two 
and two, have the same ratio ; if 
G be greater than M, H is greater 
than N ; and if equal, equal ; and 
if less, less"; and G, H are any 
equimultiples whatever of A, D, 
and M, N are any equimul- 
tiples whatever of C, F : Therefore*^, as A is 
to F. 

Next, I^t there be four magnitudes. A, B, C, D, and other 

four E, F, G, H, which, twoand two, have the 

same ratio, viz. as A is to B, so is E to F; and A. B. C. D. i 
as B to C, so F to G : and as C to D, so G to E. F. G. H. i 
H : A shall be to D, as E to H. ' 

Because A, B, C, are three magnitudes, and E, F, G, other 
three, which, taken two and two, have the same ratio ; 
by the foregoing case, A is to C, as E to G : But C is to D, 
as G is to H ; wherefore again, by the first case, A is to D, 
as E to H ; a-.id so on, whatever be the number of magnitudes. 
Therefore, if there be any number. Sec. Q^ E. D. 



"20.5. 



to C, so is D ' 5. Dcf. S. 




THE ELEMENTS 



PROP. XXIII. THEOR. 



SecN. 



15.5. 



bll.5. 



4.5. 



*21.5. 



J. F there be any number of magnitudes, and as 
many others, which, taken two and two, in a cross 
order, have the same ratio ; the first shall have to 
the last of the first magnitudes the same ratio which 
the first of the others has to the last. N. B. T/iis 
u- usually cited hy the words, "c.r cequali in pi^o- 
portione perturhata ;" or, *' ex (Zquo perturbato.''^ 

First, Let there be three magnitudes A, B, C, and other 
three D, K, F, which, taken two and two, in a cross order, 
have the same ratio, that is, such that A is to B, as E to F ; 
and as B is to C, so is D to E: A is to C, as D to F. 

Take of A, B, D any equimultiples whatever G, H, K ; and 
of C, E, F any equimultiples whatever L, M, N : And because 
G, H are equimultiples of A, B, 
and that magnitudes have the 
same ratio which their equimul- 
tiple have* ; as A' is to B, so is 
G to H : And for the same rea- 
son, as E is to F, so is M to N : 
But as A is to B, so is E to F i 
as therefore G is to H, so is M to 

N^ And because as B is to C, ^ ^ ^ K. M ? 
so is D to E, and that H, K are 
equimultiples of B, D, and L, M 
of C, E ; as H is to L, so is"^ K 
to M : And it has been shown 
that G is to H, as M to N : Then, 
because there are three magni- 
tudes G, H, L, and other three 
K, M, N, which have the same 
ratio taken two and two in a cross 
order ; if G be greater than L, 
K is greater than N ; and if equal, equal ; and if less, Icss^ ; 
and, G, K, are any equimultiples whatever of A, D; and L> 
N any whatever of C, F: as, therefore, A is to C, so is D to F, 

Next, 



ABC B 



f 




OF EUCLID. 

Next, Let there be four magnitudes, A, B, C, D, and other 
four E, F, G, H, which, taken two and two, 
in a cross order, have the same ratio, viz. 
A to B, as G to H ; B to C, as F to G ; and 
G to D, as E to F : A is to D, as E to H. 

Because A, B, C, are three magnitudes, and F, G, H other 
three, which taken two and two, in a cross order, have the 
same ratio ; by the first case, A is to C, as F to H : But C is 
to D, as E is to F; wherefore again, by the first case, A is 
to D, as E to H : And so on, whatever be the number of mag- 
nitudes. Therefore, if there be any number, &c. Q. E. D. 



PROP. XXIV. THEOR. 

1 F the nrst has to the second the same ratio which 
the third has to the fourth ; and the fifth to the se- 
cond, the same ratio which the sixth has to the 
fourth ; the first and fifth together shall have to the 
second, the same ratio which the third and sixth 
togetlier have to the fourth. 

Let AB the first, have to C the second, the same ratio which 
DE the third, has to F the fourth j and let iSG the fifth have 
to C the second, the same ratio which EH 
the sixth, has to F the fourth ; AG, the G 
first and fifth together, shall have to C the 
second, the same ratio' which DH, the 
third and sixth together, has to F the 
fourth. 

Because BG is to C, as EH to F ; by in- 
version, C is to BG, as F to EH : And be- 
cause, as AB is to C, so is DE to F ; and 
as C to BG, so F to EH ; ex a?quali% 
AB is to BG, as DE to EH : And be- 
cause these magnitudes are proportionals, 
they shall likewise be proportionals when 
taken jointly*" j as therefore AG is to GB, 
so is DH to HE ; but as GB to C, so is HE to F. 



B 



H 



A C D F 



SeeN*. 



'22,5. 



»I3. 



Th( 



re- 



fore ex aequalis as AG is to C, so is DH to F. .Wherefore 
ifthefirst, &c. Q. E. D. 

Cor. I. If the same hypothesis be made as in the proposi- 
tion, the excess of the first and fit'th shall be to the second, as 

the 




THE ELEMENTS 

the excess of the third and sixth to the fourth : The demon- 
stration of this is the same with that of the proposition, if di- 
vision be used instead of composition. 

Cor. 2 The proposition holds true of two ranks of magni-' 
tudes, whatever be their number, of whic"h each of the nrst 
rank has to the second magnitude the same ratio that the cor- 
responding one of the second rank has to a fourth magnitude ', 
as is manifest. 

PROP. XXV. THEOR. 



JLF four magnitudes of the same kind are prportion- 
als, the greatest and least of them together arc 
greater than the other two together. 

Let the four magnitudes AB, CD, E, F be proportionals, 
viz. AB to CD, as E to F i and let AB be the greatest of them, 
• A. & I'i. ^'^^ consequently F the leasts AB, together with F, are 
5. greater than CD, together with E. 

Take AG equal to E, and CH equal to F : Then because as 
AB is to CD, so is E to F, and that AG is equal to E, and CH 
equal to F, AB Is to CD, as AG to CH. 



And because AB the whole, is to the 
whole CD, as AG is to CH, likewise the p 
remainder GB shall be to the remainder 



"A. 5 






19.5. HD, as the whole AB is to the whole'' 
CD : But AB is greater than CD, there- 
fore"^ "GB is greater than HD : And be- 
cause AG is equal to E, and CH to F j 
AG and F together are equal to CH and 
E together. If therefore to the unequal x n rr t 
magnitudes GB, HD, of which GB is -^ ^ ^ ^ 
the greater, there be added equal magnitudes, viz. to GB the 
two AG and F, and CH and E to HD j AJB'and F together 
are greater than CD and E. Therefore, if four magnitudes, 
&c. Q. E. D. 

PROP. F. THEOR. 



S<«N. 



XxATIOS which are compounded of the same ra- 
tios, are the same with one another. 

Let 



OF EUCLID. 



143 




Let A be to B, as D to E ; and B to C, as E to F : The ni- ^-^ok v. 
tio which is compounded of the ratios of A ^-"v^^ 

to B, ana B to C, which by the definition 
of compound ratio, is the ratio of A to C,^ is 
the same with the ratio of D to F, which by 
the same definition, is compounded of the 
ratios of D to E, and E to F. 

Because there are three magnitudes A, B, C, and three 
others D, E, F, which, taken two and two, in order, have 
the same ratio ; ex aequali A is to C, as D to F*. 

Next, Let A be to B, as E to F, and B to C, as D to E ; there- 
for?, ex aquali in proportion perturbatct^^ A 
is to C, as D to F ; that is, the ratio of A to 
C, which is compounded of the ratios of A 
to B, and B to C, is the same with the ratio 
of D to F, which is compounded of the ra- 
tios of D to E, and E to F : And in like manner the proposi- 
tion may be demonstrated, whatever be the number of ratios 
in either case. 



'22 5. 
b23. 5. 



A. B. C. 
D. E. F. 



PROP. G. THEQR. 



IF several ratios be the same with several ratios, c v 
each to each ; the ratio which is compounded of 
ratios which are the same with the first ratios, each 
to each, is the same with the ratio compounded 
of ratios w hich are the same with the other ratios, 
each to each. 

Let A be to B, as E to F ; and C to D, as G to H : A.nd let 
A be to B, as K CO L ; and C to D, as L to M : Then the ra, 

tio of K to M, by the definition - — . 

of compound ratio, is compound- 1 A. B C D K L M 



E. F. G. H. N. O. P. 



ed of the ratios of K to L, and 
L to M, which are the same with 
the ratios of A to B, and C to D : 
And as E to F, so let N be to O ; and as G to H, so let O be 
to P J then the j-aiio of N" to P, is compounded of the ratios 
of N to O, and O to P, which are the same with the ratios of 
E to F, and G to H : And it is to be shown that the ratio of 
K to M, is the same with the ratio of N to P, or that K is to 
M, asNtoP. 

Because K is to L, as (A to B, that is, as E to F, that is, as 
N toOi and as L to M) so is (C to D, and so is G to H, 

and 




SeeN. 



THE ELEMENTS 

and so is'O to P:) Ex xquali=' K is to M, as N to P. Thetc- 
»22. 5. fore, if several ratios, &c. Q. E. D. 

PROP. H. THEOR. 

IF a ratio compounded of several ratios be the 
same with a ratio compounded of any other ratios, 
and if one of the first ratios, or a ratio compoun- 
ded of any of the first, be the same with one of the 
last ratios, or with the ratio compounded of any 
of the last ; then the ratio compounded of the re- 
maining ratios of the firsts or the remaining ratio 
of the first, if but one remain, is tlie same with the 
ratio compounded of those remaining of the last, or 
with the remaining ratio of the last. 

Let the first ratios be those of A to B, B to G> C to D, 

D to E,^and E to F ; and let the other ratios be those of G to 

H, H to K, K to L, and L to M ; Also, let the ratio of A to 

o?coml"°" F, which is compounded of^ the first 



A. B. C. D. E. F. 
G. H. K. L. M. 



pounded ratios,be the same with theratioof G 

"''°* to M, which is corapounded of the 
other ratios: And besides, let the ra- 
tio of A to D, which is compounded of the ratios of A to B, 
B to C, C to D, be the same with the ratio of G to K, which 
is compounded of the ratios of G to H, and H to K : Then 
the ratio compounded of the remaining first ratios, to wit, of 
the ratips of D to E, and E to F, which compounded ratio is 
the ratio of D to F, is the same with the ratio of K to M, 
which is compounded of the remaining ratios of K to L, and 
L to M of the other ratios. 

Because^ by the hypothesis, A is to D, as G to K, by in- 

b B 5. version'', D is to A, as K to G ; and as A is to F, so is G to 

' 22. 5. M J therefore^ ex aequali, D is to F, as K to M. If there- 
fore a ratio which is, &c, Q^ E. D. 



O^ EUCLia I4S 

BcoxV. 

PROP. K. THEOR. 

J.F there be any number of ratios, and any number see n. 
of other ratios such, that the ratio compounded of 
ratios which are the same with the first ratios, each to 
each, is the same with the ratio compounded of ratios 
"which are the same, each to each, with the last ratios; 
and if cne of the first ratios, or the ratio which is 
' compounded of ratios which are the same with several 
of the fii-st ratios, each to each, be the same with one 
of the last ratios, or with the ratio compounded of ra- 
tios which are the same, each to each, with several of 
the last ratios : Then the ratio compounded of ratios 
which arc the same with -the remaining ratios of the 
first, each to each, or the remaining ratio of the first, 
if but one remain ; is the same with the ratio com- 
pounded of ratios which are the same with those re- 
maining of the last, each to each, or with the re- 
maining ratio of the last. 

Let the ratios of A to B, C to D, E to F, be the first ratios ; 
and the ratios of G to H, K to L, .\1 to N, O to P, Q to R, 
be the other ratios : And let A be to B, as S to T ; and C to 
D, as T to V, and E to F, as V to X : Therefore, by the deii- 
nition of compound ratio, the ratio of S to X is compounded 



h, k, 1. 
A, Bi C, D; E, F. S, T, V. X. 

G,H;K, LjM,NjO,PiQjR. Y, Z,a, b, c,d. 
e, f, g. . m, n, o, p. 



t)f the ratios of S to T, T to V, and V to X, which are the 
Isame with the ratios of A to B, C to D, E to F, each' to each : 
I Also, as G to H, so let Y be to Z ; and K to L, as Z to a ; M 
Ito N, as a to b, O to P, as b to c j and Q^to R, as c to d : 
ITherefore, by the same definition, the ratio of Y to d is com- 
Ipounded of the ratios of Y to Z, Z to a, a to b, b to c, and 
I L c to 



146 



THE ELEMENTS OF EUCLID. 



BookV^ q to d, which are the same, each to each, with the ratios of G 
''"^^'^^ to H, K to L, M to N, O to P, ancl Qjo R : Therefore, by the 
hypothesis, S is to X, as Y to d ; Also, let the ratio of A to B, 
, that is, the ratio of S ,to T, which is one of the first ratios, be 
the same with the ratio of e to g, which is compounded of the 
ratios of e to f, and f to s;, which, by the hypothesis, are the 
same with the ratios of G to H, and K to^ L,two of the other 
ratios j and let the ratio of h to 1 be that v/hich is compounded 
of the ratios of h to k, and k to 1, which are the same with the 
remaining first ratios, viz. of C to D, and E to F 5 also, let 
the ratio of m to p, be that which is compounded of the ratios 
of m to n, n to o, and o to p, which are the same, each to each, 
with the remaining other ratios, viz. of iM to N, O to P, and 
Q to R : Then the ratio of h to 1 is the same with the ratio 
of m to p, or h is to 1, as m to p. 



U.5. 



h, k, i. 




A, B ; C, D i E, F. 


S,T,V,X. 


G, H; K, L; M, N J 0, P; Q, P« 


Y, Z, a, b. c, d. 


e, f, g. m, n, 0, p. 


- 



Because c is,' to f, as (G to H, that I?, as) Y to Z j and f is 
to g, as (K to L, that is, as) Z to a ; therefore, ex icquali, e is 
to g, as Y to a : And by the hypothesis, A is to B, that is, 
S to X, as e to g ; wherefore S is to T, as Y to a ; and, by 
inversion, T is to S, as a to Y ; and S is to X, as Y to d ; 
therefore, ex sequali, T is to X, as a to d : Also, because h is 
to k as (C to D, that is, as) T to V j and k is to 1, as (E to 
F, that is, as) V to X ; therefore, excequali, h is to 1, ?.s T 
to X : In like manner, it may be demonstrated, that m is to 
p, as a to d : And it has been shown, that T is to X, as a to 
d ; therefore" h is to 1, as m to p. Q. E. D. 

The propositions G and K are usually, for the sake of bre- 
vity, expressed in the same terms with propositions F and H: 
And therefore it was proper to show the true meaning of them 
when they are so expressed ; especially smce they are very 
frequently made use of by geometers. 



[ 147 ] 



THE 



ELEMENTS 



EUCLID. 



BOOK VI. 
DEFINITIONS. 



I. 



Similar re£tilineal figures 
are those which have their 
several angles equal, each to 
each, and the sides about the 
equal angles proportionals. 




Book VI. 



^ 



II. 



"Reciprocal figures, viz. triangles and parallelograms, are See N. 

" such as have their sides about two of their angles propor- 

*' tionals in such manner: that a side of the first figure is to 

*' a side of the other, as the remaining side of this other is 

** to the remaining side of the first." 
III. 
A straight line is said to be cut in extreme and mean ratio, 

when the whole is to the greater segment, as the greater 

segment is to the less.^ 

IV. 
The altitude of any figure is the straight 

line drawn from its vertex perpendicular 

to the base. 





THE ELEMENTS 



PROP. I. THEOR. 



seeN. j^ RiANGLES and parallelograms of thc saiiic al- 
titude are one to another as their bases. 

Let the triangles ABC, ACD, and the parallelograms EC, 

CF have the same ali'tude, viz. the perpendicular drawn from 

the point A to BD : Then, as the base BC, is to the base CD, 

so is the triangle ABC to the triangle ACD, and the paral- 

'lelogram EC to the parallelogram CF. 

Produce BD both ways to the points H, L, and take any 
number otst;raight Imes BG, GH, each equal to the base BC; 
andDFC, KL, any number of them, each equal to the base CD ; 
and join AG, AH, AK, AL : Then, because CB, BG, GH 

'\ 1. are all equal, the triangles AHG, AGB', ABC are all equal'T 
Therefore, whatever multiple the base HC is of the base BC, 
the same multiple is the. triangle AHC of the triangle ABC : 
For the same reason, whatever multiple the base LC is of the 
base CD, the same multi- 
ple is the triangle ALC E A F 
of the triangle. ADC: 
And if the base HC be 
equal to the base CL, the 
triangle AHC is also 
equal to the triangle 
ALO^; and if the base 
HC be greater than the 
base CL, likewise the 
triangle AHC is greater than the triangle ALC; and if,les3-, 
less : Therefore, since there are four magnitudes, viz. the two 
bases BC, CD, and the two triangles ABC, ACD ; and of the 
base BC and the triangle ABC, the first and third, any equi- 
multiples whatever have been taken, viz. the base HC and tri- 
angle AHC; and of the base CD and triangle ACD, the se.^ 
cond and fourth, have been taken any equimultiples whatever, 
viz. the base CL and triangle ALC ; and that it has been shown, 
that, if the base HC be greater than the base CL, the triangle 
AHC is greater than the triangle ALC ; and if equal, equal -^ 

bA d«f. 5. ^"d if less, less : Therefore^ as the base BC is to the base CD, 
so is the triangle ABC to the triangle ACD. 

And because the parallelogram CE is double of the triangle 

ABC, 




OF EUCLID. 149 

ABO and the parallelogran; CF double of the triangle ACD, ^''^• 
and that magnitudes have i.he same ratio which their equimul-c 41. 1. 
tiples hav^e ; as the triangle ABC is to the triangle ACD, so Mj. 5. 
is the parulicio^ram EC to the parallelogram CF : And because 
it has been s.io\A'n, that, as the base BC is to the base CD, so 
is the triangle ABC to t.^e triangle ACD j and as the triangle 
ABC is to the triangle A_D, so is the parallelogram EC to 
the para.lelogram CF ; therefore, as the base BC is to tiie base 
CD, so is^ the parallelogram EC to the parallelogram CF. en.5. 
Wherefore triangles, &c. Q. E. D. 

Cor. From this it is plain, that triangles and parallelo- 
grams that have equal altitudes are one to another as their bases. 

Let the.r figures be placed so as to have their bases in the 
same straight line ; and having drawn perpendiculars from 
the vertices of the triangLs to the bases, the straight line 
which joins the vertices js parallel to that in which their 
bases arc*", because the perpendiculars are both equal and pa- '33.1. 
rallel to one another. Then, if the same construction be made 
as in the proposition, the demonstration will be the same. 

PROP. II, THEOR. 

^ F a straight line be drawn parallel to one of the see n. 
sides of a triangle, it shall cut the other sides, or 
these produced, proportionally : And if the sides, 
or the sides produced, be cut proportionally, the 
straight line which joins the points of section shall 
be parallel to the remaining side of tlie triangle. 

Let DE be drawn parallel to BC, one of the sides of the 
triangle ABC : BD is tt^ DA, as CE to EA. 

Join BE, CD ; then the triangle BDE is equal to the tri- 
angle CDE'*, because they are on the same base DE, and be- »37. i. 
twcen the same parallels DE, BC : ADE is another triangle, 
and equal magnitud s have the same, the same ratio'' j there- "T-S. 
iu^e^ as the triangle BDE to the triangle ADE,-so is the tri- 
angle CDE to the triangle ADE , but as the triangleBDE to 
the triangle ADE, so is<: BD to DA, because having the same c 1. ,-. 
iltitude, viz. t;.e perpendicular drawn from the point E to AB, 
ley are to one anoj^er as their basest and for the same reason, 

L 3 as 



150 



THE ELEMENTS 



BooK^. as tiie triangle CDE to the triangle ADE, so is CE to EA : 
■uTs, Therefore, as BD to DA, so is CE to EA**. 

Next, Let the sides AB, AC, of the triangle ABC, or these 

A. A Ik. D 



•1.6, 



f9.5. 



«3a.i. 



3eeN. 




C B E B C 

produced, be cut proportionally in the point D, E, that is, so 
that BD be to DA as CE to EA, and join DE ; DE is paral- 
lel to BC. 

The same construction being made, Because as BD to DA, 
so is CE to EA ; and as BD to DA, so is the triangle BDE 
to the triangle ADE^ ; and as CE to EA, so is the triangle 
CDE to the triangle ADE ; therefore the triangle BDE is to 
the triangle ADE, as the triangle CDE to the triangle ADE ; 
that is, the triangles BDE, CDE have the same ratio to the 
triangle ADE ; and thercforc*^ the triangle BDE is equal to 
the triangle CDE: and they are on the same base DE ; but 
equal triangles on the same base are between the same paral- 
lels s ; therefore DE is parallel to BC. Wherefore, if a straight 
line, &c. 0. E. D. 

PROP. in. THEOR. 

xF the angle of a triangle be divided into two eqnril 
angles, by a straight line M'hich also cuts the bas( , 
the segments of the base shall have the same ratio 
whicli the other sides of the triangle liave to one 
another : And if the segments of tlie base have tlic 
same ratio which the other sides of the triangle 
have to one another, the straight line drawn fniin 
the vertex to the point of section, divides the vci- 
ticle angle into two equal angles. 

Let the angle BAC of any triangle ABC be divided into twi 
equal angles by the straight line AD : BD is to DC, as B/ 
to AC. 

Throuul 



OF EUCLID. 



151 




^o.l. 



Through the point C draw CE parallel to DA, and let BA Book VI. 
produced meet CE in E. Because the straight line AC meets ^^^^"^^ 
the parallels AD, EC the angle ACE is equal to the alternate '^ " ' 
angle CAD*": But CAD, by the hypothesis, is equal to the ^ 
an^ie BAD; wherefore BAD is equal to the angle ACE. " ' 
Again, because the straight line 

BAE meets the parallels AD, K 

EC, the outward angle BAD 
is equal to the Inward and ^ 

opposite angle AEC : but the 
angle ACE has been proved 
equal to the angle BAD ; there- 
fore also ACE is equal to the 
angle AEC, and consequently iT' 
the side AE is equal to the 

side*= AC : And because AD is drawn parallel to one of the 
sides of the triangle BCE, viz. to EC, BD is to DC, as BA 
to AE"^, but AE is equal to AC j therefore, as BD, DC, so is <* 2. 6. 
BAtoAC^ 

Let now BD be to DC, as BA to AC, and join AD ; the * "' ^' 
angle BAC is divided into two equal angles by the straight 
line AD. 

The same construction being made ; because, as BD to DC, 
so is BA to AC i and as BD to DC, so is BA to AE"^, because 
AD is parallel to EC ; therefore BA is to AC, as BA to AE^ : '"• 5. 
Consequently AC is equal to AEs, and the angle AEC is there- * 9. 5. 
fore equal to the angle ACE^ : But the angle AEC is equal to " 5.1. 
the outward and opposite angle BAD ; and the angle ACE 
is equal to the alternate angle CAD**: Wherefore also the an- 
gle BAD is equal to the angle CAD : Therefore the angle 
BAC is cut into two equal angles by the straight line AD, 
Therefore, if the angle, &c. Q. E. D. 



L4 



152 THE ELEMENTS 

Book VL, 

PROP, A. THEOR. 

If the outward angle of a triangle made by pro-, 
ducing one of its sides, be divided into two equal 
angles, by a straight line which also cuts the base 
produced ; the segments between the dividing line 
and the extremities of the base have the same ratio 
which the other sides of the triangle have to one 
another : And if the segments of the base produced, 
have the same ratio which the other sides of the tri- 
angle have, the straight hne drawn from the ver- 
tex to the point of section divides the outward an- 
gle of the triangle into two equal angles. 

Let the outward angle CAE of any triangle ABC bedivided 

into two 'equal angk's by the straight line AD which meets 

the base produced in D : BD is to DC, as BA to AC. 
?3I. 1. Through CdrawCF parallel to i\D*i and because the straight 

line AC meets the paiallels AD, tC, the angle ACF is equal 
"29. 1. to the alternate angle CAD^ ; But CAD is equal to the angle 
« Hyp. DAE^; therefore' also DAE is equal to the angle ACF. Again, 

because the straight line FAE meets the parallels AD, FC, the 

outward angle DAE is 

equal to the inward and op- j^ 

posite angle CKA: But the 

angle ACF has been proved 

eq^ual to the angle DAE; 

therefore also the angle 

ACF is equal to the angle 

CP'A, and consequently the 

side AF IS equal to the side 
«5. 1. ■ AC** : And because AD is parallel to FC, a side of the -tri-. 
«2. 6. angle BCF, BD is to DC, as BA to AFs but AF is equal to 

AC ; as therefore BD is to DC, so is BA to AC. 

Let n )W BD be to DC, as BA to AC, and join AD ; the 

angle CAD is equal to the angle DAE. 

T e same construction being made, because BD is to DC, 
ni. 5. as B.\ to AC ; and that BD is also to DC, as BA, to AF= ; 
e 9. 5. therefore BA is to AC, as BA to AF*^: wherefore AC is equal 
b 5^ 1. to AF?, and the angle AFC equal*' to the angle ACF : But 

the 




OF EUCLID. 153 

the angle AFC is equal to the outward angle EAD, and the ^°°^ ^'^' 
angle ACF to the alternate angle CAD ; therefore also EAD ^^'^''"*^ 
is equal to the angle CAD. Wherefore, if the outward, &c. 
Q.E. D. 

PROP. IV. THEOR. 

X HE sides about the equal angles of equiangular 
triangles are proportionals ; and those which are op- 
posite to the equal angles are homologous sides, that 
IS, are the antecedents or consequents of the rajtios. 

Let ABC, DCE be equiangular triangles, having the angle 
ABC equal to the angle DCE, and the angle ACB to the an- 
gle DEC, and consequently^ the angle BAC equal to the angle » 32. 1, 
CDE. The sides about the equal angles of the triangles 
ABC, DCE are proportionals ; and those are the homologous 
sides which are opposite to the equal angles. 

Let the triangle DCE be placed, so that its side CE may be 
contiguous to BC, and in the same straight line with it : And 
because the angles ABC, ACB are together less than two 
rightangies\ABC,andDEC,which ''^''- ^* 

is equal to ACB, are also less than F\ 
two right angles; wherefore BA, I 

ED produced shall meet<^; let them A \^ 'la. Ax. 1, 

be produced and meet in the point 
F ; and because the angle ABC is 
equal to the angie DCE, BF is pa- 
rallel'' to CD. Again, because the p. n^^ >. ^,,3 j 
angle ACB is equal to the angle * 
DEC, AC is parallel to FE-^: 

Therefore FACD is a parallelogram ; and consequently AF is 
equal to CD, and AC to FD'=: And because AC "is parallel «5i. I, 
to FE, one of the sides of the triangle FBE, BA is to AF, as 
BC to CE' : But AF is equal to CD ; thereforee, as BA to ^l- 6. 
CD, so IS BC to CE ; and alternately, as AB to BC, so is DC * '^' ^• 
to CE : Again, because CD is parallel to BF, as BC to CE, 
so is FD to DE^ ; but FD is equal to AC; therefore, as BC ' 
to CE, so is AC to DE •. And alternately, as BC to CA, so 
CE to DE : rherefore, because it has been proved that AB is 
to BC, as DC to CE, and as BC to CA, so CE to ED, ex 
equallS BA is to AC as CD to DE. Therefore the sides, "^ 22- J". 
&c. Q.E. D. 




154 THE ELEMENTS 

Book Vr, 



PROP. V. THEOR. 



] 



'23. 1. 



F the sides of two triangles, about each of their 
singles, be proportionals, the triangles shall be equi- 
angular, and have their equal angles opposite to the 
homologous sides. 

Let the triangles ABC, DEFhave their sides proportionals, 
so that A B is to BC, as DE to EF ; and BC to CA, as EF to 
FD ; and consequently, ex aequali, BA to AC, as ED to DF ; 
the triangle ABC is equiangular to the triangle DEE, and 
their equal angles are opposite to the homologous sides, viz. 
the angle ABC equal to the angle DEE, and BCA to EFD, 
and also BAG to EDF. 

At the points E, F, in the straight line EF, makes^ the angle 
FEG equal to the angle ABC, and the angle EFG equal to 
BCA ; wherefore the remain- 
ing angle BAC is equal to the 
^' remaining angle EGF'=, and 
the triangle ABC is there- 
fore equiangular to the tri- 
angle GEF ; and consequently 
they have their -sides opposite 
to ^Ke- equal angles propor- 
'*-fi- tionals'=. Wherefore, . as AB 
to BC, so is GE to EF ; but 
«' 11.5, as AB to BC, so is DE to EF ; therefore as DE to EF, so*" 
GE to EF : Therefore DE and GE have the same ratio to 
EF, and consequently are equal"" : For the same reason DF is, 
equal to FG: And because, in the triangk^s DEE, GEF, DE 
is equal to EG, and EF common, the two sides DE, EF, are 
equal to the two GE, EF, and the base DF is equal to the 
base GF j therefore the angle DEE is equaK to the angle GEF, 
and the other angles to the other angles which are subtended 
by the equal sidess. Wherefore the angle DEE is equal to 
the angle CFE, and EDF to EGF : And because the angle 
DEF is equal to the angle GEF, and GEF to the angle ABC ; 
therefore the angle ABC is equal to the angle DEI" : For the 
same reason, the angle ACB is equal to the angie DEE, and 
the angle at A, to tlie angle at D. Therefore the triangle 
ABC is equiangular to the triangle DEF. Wherefore, if 
the sides, &c. Q^ E. D. 




•9. 



V. :j. 



e4. 1. 




OF EUCLID. 



PROP. VI. THEOR. 

JL F two triangles have one angle of the one equal 
to one angle of the other, and the sides about the 
equal angles proportionals, the triangles shall be 
equiangular, and shall have those angles equal which 
are opposite to the' homologous sides. 

Let the triangles ABG, DEF have the angle BAC in the 
one equal to the angle EOF in the other, and the sides about 
those angles proportionals ; that is, B A to AC, as ED to DF ; 
the triangles ABC, DEF are equiangular, and have the an^Ie 
ABC equal to the augle DEF, and ACB to DFE. 

At the points D, F, in the straight line DF, make* the an- '^^c- 
gl? FDG equal to either of the angles BAC, EDF; and the 
angle DFG equal to the an- 
gle ACB : Wherefore the 
remaining angle at B is 
equal to the remaining one 
at G'', and consequently 
the triangle ABC is equi- 
angular to the triangle 
DGF ; and therefore as 
BA to AC, so is^ GD 
to DF J but by the hypo- 
thesis, as BA to AC, so is ED to DF; as therefore ED to 
DF, so is ^ GD to DF ; wherefore ED is equal '^ to DG; and 11^'^^' 
Dp' is common to the two triangles EDF, GDF : Therefore 
the two sides ED, DF are equal to the two sides GD, DF; 
and the angle EDF is equal to the angle GDF ; wherefore 
the base EF is equal to the base FG'', and the triangle EDF to "^^ ^• 
the triangle GDF, and the remaining angles to the remaining 
angles, each to each, which are subtended by the equal sides : 
1 herefore the angle DFG is equal to the angle DFE, and the 
angle at G to the angle at E : But the angle DFG is equal to 
the angle ACB ; therefore the angle ACB is equal to the an- 
gle DFE : A^nd the angle BAC is equal to the angle EDFs ; ' Hj-p. 
wherefore also the remaining angle at B is equal to the re- 
maining angle at E. Therefore the triangle ABC is equi- 
angular to the triangle DEF. Wherefore, if two triangles, 
kc. Q. E. D. 




'3*. 1, 



♦. G. 



J56 



Book VI. 



THE ELEMENTS 



PROP. VII. THEOR. 



SeeN. 



»23. I. 



«'32. 1. 

••i. 6. 

'■=11.5. 
« 9. 5. 
^5. 1. 

<13. I. 



iF two triangle ^ have one angle of the one equal 
to one angle of the other, and the sides ahout two 
other angles proportionals, then, if each of the re- 
maining angles be either less, or not less, than a 
right angle ; or if one of them be a right angle : 
The triangle shall be equiangular, and have those 
angles equal about which the sides are proportionals 

Let the two triangles ABC, DEF have one angle in the one 
equal to one angle in the other, viz. the angle BAG to the an- 
gle EDF, and the sides about two other angles ABC, DEF 
proportionals, so that AB is to BC, as DE to EF ; and, in the 
iirst case, let each of the remaining angles at C, F be less than 
a right angle. The triangle ABC is equiangular to the trian- 
gle OEF, viz. the angle ABC is equal to the angle DEF, and 
the remaining angle at C to the remaining angle at F. 

For if the angles ABC, DEF he not equal, one of them is 
greater than the other : Let ABC be the greater, and at the 
point B^ in the straight line 
AB, make the angle ABG 
equal to the angle ^ DEF ; 
And because the angle at A 
is eq^ial to- the angle at D, 
and the angle ABGr to the 
angle DEF ; the remaining 
angle AGB is equal "^ to the 
remaining angle DFE : Therefore the triangle ABG is equi- 
angular to t"he triangle DEF ; wherefore '^ as AB is to BG, so 
is DE to EF ; but as D hi CO EF, se, by hypothesis, is A B to 
BC ; therefore as AB to BC, so is AB to BG"* : and because 
AB has the same ratio to each of the lines 'BC, BG ; BC is 
equal = to BG; and therefore the angle BGC is equal to the an- 
gle BCG*^ : But the angle BCG is, by hypothesis, less than a 
right angle ; therefore also the angb BGC is less than a right 
angle, and the adjacent angle AGB must be greater than a 
right angle E. But it was proved that the angle GB is equal 
to the angle at F ; therefore the angle at F is greater than a 
right angle : But, by the hypothesis, it is less than a right 

I angle i 




OF EUCLID. 



^51 




\T 




angle ; which is ahsurd. Therefore the angles ABC, DEF l^oo^'Vi. 
are not unequal, that is, they are equal : And the angls at A 'S ^"^^*^ 
c;qual to the angle at D j wherefore the remaining angle at C 
Ts equal to the remaining angle at h : Therefore the triangle 
ABC is equiangular to the triangle DEF. 

Next, Let each of the angles at C, F be not less than a 
right angle : The triangle ABC is also in this case equiangu- 
lar to the triangle DEF. 

The same construction 
being made, it may be j^ 

proved in like manner that 
BC is equal to BG, and 
the angle at C equal to the ^ /A- 

angle BGC: But the angle t>^ 
at C is not less than a right C 

angle ; therefore the angle 

BGC is not less than a right angle : Wherefore two angles of 
the triangle BGC, are together not less than two right angles, 
which is impossible'' ; and therefore the triangle ABC may be " ^'^' ^• 
proved to be equiangular to the triangle DEF, as in the first 
case. 

Lastly, Let one of the angles atC, F, viz. the angle at C, 
be a right angle; in this case likewise the triangle ABC is 
equiangular to the triangle DEF. 

P'or if they be not equian- 
gular, make, at the point B 
of the straight line AB, the 
angle ABG equal to the an- 
gle DEF ; then it may be 
proved, as in the first case, 
that BG is equal to BC : But 
the angle BCG is a right an- 
gle,therefore^ the angle BGC 
is also a right angl.e ; whence 
two of the angles of the tri- 
angle BGC are together not 
less than two right angles, p 
which isimpessible*": There- 
fore the triangle ABC is equi- 
angular to the triangle DEF. Wherefore, if the two triangles, 
&c, Q. E. D. "^ 




15? 



Book VI. 



THE ELEMENTS 



SeeN, 



n2, 1. 



"•4. 6. 
' 1. clef. 6. 



PROP. VIII. THEOR. 

jN a right angled triangle, if a perpendicular be 
drawn from the right angle to the base ; the trian- 
gles each side of it are similar to the whole triangle, 
and to one another. 

Let ABC be a right angled triangle, having the right angle 
BAG ; and from the point A let AD be drawn perpendicular 
to the base BC : The triangles ABD, ADC are similar to the 
whole triangle ABC, and to one another. 

Because the angle B AC is equal to the angle ADB, each of 
them being a right angle, and that the angle at B is common 
to the two triangles, ABC, 
ABD : the remaining angle 
ACB is equal to the remaining 
angle BAD =^ : Therefore the 
triangle ABC is equiangular 
to the triangle ABD, ^nd the 
sides about their equal angles 
are proportionals^'' ; wherefore 
the triangles are similar'^: In • 

the like manner it may be demonstrated, that the triangle 
ADC is equiangular and similar to the triangle ABC : And 
the triangles ABD, ACD, being both equiangular and similar 
to ABC, are equiangular and similar to each other. There- 
fore, in a right angled. Sec. Q. E. D. 

CoR. From this it is manifest that the perpendicular drawn 
from the right angle of a right angled triangle to the base, is a 
mean proportional between the segments of the base : And 
also, that each of the sides is a mean proportional between the 
base, and its segment adjacent to that side : Because in the 
triangles BDA, ADC, BD is to DA, as DA to DC' ; and in 
the' triangles ABC, DBA, BC is to BA, as BA to BD"; and 
in the triangles ABC, ACD, BC is to C A, as CA to CD". 




OF EUCLID. 



Book Vf. 



PROP. IX. PROB. 

Jr RO^I a given straight line to cut oiF any partsecN. 
required. 

Let AB be the given straight line ; it is required to cutofF 
any part from it. 

From the point A draw a straight line AC, making any an- 
gle with AB ; and in AC take any point D, and take AC the 
same multiple of AD, that AB is of the part 
which is to be* cut off from it; join BC, 
and draw DE parallel to it : Then AE is 
the part required to be cut off. 

Because ED is parallel to one of the sides 
of the triangle ABC, viz. to BC, as CD is 
to DA, so is * BE to EA : and, by compo- 
sition^, CA is to AD, as BA to AE : But 
CA is a multiple of AD; therefore'^ BA 
is the same multiple of AE : Whatever part 
therefore AT) is of x\C, AE is the same 
part of AB: Wherefore, from the straight 
line AB the part required is cut ofF. .Which was to be done. 




» 13. 3. 



PROP. X. PROB. 

1 O divide a given straight line similarly to a given 
divided straight line, that is, into parts that shall 
have the same ratios to one another which the parts 
of the divided given straight line have. 

Let AB be the straight line given to be divided, and AC the 
divided line : it is required to divide AB similarly to AC. 

Let AC be divided in the points D, E ; and let AB, AC be 
placed so as to contain any angle, and join BC, and through 
the points D, E, draw^ DF, EG parallels to it; and through 131. 1. 
Ddraw DHK. parallel to AB : Therefore each of the figures 
FH, HB, is a parallelogram ; wherefore DH is equaPto FG, "3*. 1. 

and 



2.6. 



i6o T H fe E L E M E N T S 

^okVL a„jj HK to GB : And because HE is 
parallel to KC, one of the sides of 
the triangle D:vC, as CE to ED, 
so is "^ KH to HD : But KH is equal 
to BG, and HD to GF ; therefore, as 
CE to ED, 'SO is BG to GF : Again, 
because FD is parallel to EG, one of 
the sides of the triangle AGE, as ED 
to DA, so is GF to FA : But it has 
been proved that CE is to ED, as 

BG to GF ; and as ilD to DA, so GF to FA : Therefore, 
the given straight line AB is divided similarly to AC. Which 
was to be done. 




PROP. XI. PROB. 



•2. 6. 



X O find a tliird proportional to two given straight 
lines. 

Let AB, AC be the two given straight lines, and let them 
be placed so as to contain any angle ; it is 
required to find a third proportional to AB j 
AC. 

Produce AB, AC, to the points D, E ; 
and make BD equal to AC ; and having 
joined BC, through D, draw DE parallel to 
it.» 

Because BC is parallel to DE, a side of 

the triangle ADE, AB is '' to BD, as AC ry ^ 

to CE : But BD is equal to AC ; as there- ^ 
fore AB to AC, so is AC to CE. Wherefore,' to the two 
given straight lines AB, AC a third proportional CE is found. 
Which was to be done. 




PROP. XII. PROB. 

1 O find a fourth proportional to three given 
straight lines. 

Let A, B, C be the three given straight lines j it is re- 
quired to find a fourth proportional to A, B, C. 

Take 



OF EUCLID 



i6i 



• 31. !. 



Take two straight lines DE, DF, containing any angle ^^^i 
EDF ; and upon these make 
DG equal to A, GE equal to 
B, and DH equal to C 5 and 
having joined GH, draw EF 
parallel » to it through the 
point E : And because GH 
is parallel to EF, one of the 
sides of the triangle DEF, 
DG is to GE, as DH to 

HF'' ; but DG is equal to j<^ ft' •> 2. 6. 

A, GE to B, and DH to 

C; therefore, as A is to B, so isC to HF. Wherefore to the 
three given straight lines A, B, C, a fourth proportional HF 
is found. Which was to be done. 




PROP. XIII. PROB. 



JL O find a mean proportional between two given 
straight lines. 

Let AB, BC be the two given straight lines ; it is required 
to find a mean proportional between them. 

Place AB, BC in a straight line, and upon AC describe the 
semicircle ADC, and from the 
point B draw ^ BD at right an- 
gles to AC, and join AD, DC. 

Because the angle ADC in a 
semicircle is a right angle°, and 
because in the right angled tri- 
angle ADC, BD is drawn from 
the right angle perpendicular to 
the base, DB is a mean propor- 
tional between AB, BC the segments of the base= : Therefore c cox 8 6. 
between the two given straight lines AB, BC, a mean propor« 
tional DB is found. Which was to be done. 

M 




» 11. 1. 



" 31. 3. 



x62 THE ELEMENTS 

BookTI. ^ 

PROP. XIV. THEOR. 

JiQUAL parallelograms, which have one angle of 
the one equal to one angle of the other, have their 
sides about the equal angles reciprocally propor- 
tional : And parallelograms that have one angle of 
the one equal to one angle of the other, and their 
sides about the equal angles reciprocally propor- 
tional, are equal to one another. 

Let AB, BC be equal parallelograms, which have the an- 
gles at B equal, and let the sides DB, BE be placed in the 
same straight line ^ wherefore also FB, BG are in one straight 

'i 14. 1. line*. The sides of the parallelograms AB, BC about the 
equal angles, are reciprocally proportional j that is, DB is to 
BE, asGB.toBF. 

Complete the parallelogram FE ; and because the parallelo- 
gram AB is equal to BC, and 
that FE is another parallelo- 
gram, AB is to FE, as BC to 

^rs. p£b. g^t^s ABtoFE, so is 

' ^- 6. the base DB to BE^ ; and as 
BC to FE, so is the base of GB 
toBF; therefore, as DB to BE, 

* 21- 5. so is GB to BF'i. Wherefore, 
the 6ides of the parallelograms 
AB, BC about their equal an- 
gles are reciprocally proportional. 

But, let the sides about the equal angles be reciprocally pro- 
portional, viz. as DB to BE, so GB to BF; the parellelo- 
gram AB is equal to the parallelogram BC. 

Because, as DB to BE, so is GB to BF ; and as DB to 
BE, so is the parallelogram AB to the parallelogram FE ; and 
as GB to BF, so is the parallelogram BC to the parallelogram 
FE i therefore as AB to FE, so BC to FE'' : Wherefore the 
parallelogram AB is cqual<^ to^he parallelogram BC. There- 
fore equal parallelograms, &c. Q.ED. 




« 9. 5. 




OF EUCLID. 



PROP. XV. THEOR. 

il^QUAL triangles which have one angle of the 
other equal to one angle of the other, have their 
sides ahout the equal angles reciprocally proportion- 
al : And triangles which have one angle in the one 
equal to one angle in the other, and their sides ahout 
the equal angles reciprocally proportional, are equal 
to one another. 

Let ABC, ADE be equal triangles, which have the angle 
BAG equal to the angle DAE ; the sides about the equal an- 
gles of the triangles are reciprocally proportiogal j that is, CA 
is to AD, as EA to AB, 

Let the triangles be placed so, that their sides, CA, AD be 
in one straight line ; wherefore also EA and AB are in one 
straight line* : and join BD. Because the triangle ABCis*!-*. i. 
equal to the triangle ADE, and 
that ABD is another triangle ; 
therefore as the .triangle CAB is 
to the triangle BAD, so is triangle 
EAD to triangle DAB** : But as 
triangle CAB to triangle BAD, 

ISO is the base CA to AD*^ j and as 
triangle EAD to triangle DAB, 
so is the base EA to AB= : as 
'therefore CA to AD, so is EA to 
AB'' ; wherefore the sides of the triangles ABC, ADE about " "• 5. 
the equal angles are reciprocally proportional. 

But let the sides of the triangles ABC, ADE about the equal 
angles be reciprocally proportional, viz. CA to AD, asEA to 
AB ; the triangle ABC is equal to the triangle ADE. 

Having joined BD as before j because, as CA to AD, so is 
EA to AB J and as CA to AD, so is triangle ABC to triangle 
BAD-; and as EA to AB, so is triangle EAD to triangle 
^D' ; therefore'* as triangle BAC to triangle BAD, so is^ 
■ :igle EAD to triangle BAD; that is, the triangles BAC, 
vD have the same ratio to the triangle BAD: Wherefore 
J triangle ABC is equal' to the triangle ADE. Therefore « 9. 5. 
-al triangles, &c. 6. E. D. 
M 2 




164 
ioot yj. 



THE ELEMENTS 



PROP. XVI. THEOR. 



II. I. 



7. 5. 



14. 6. 



XF four straight lines be proportionals, the rectan- 
gle contained by the extremes is equal to the rectan- 
gle contained by the means : And if the rectangle , 
contained by the extremes be equal to the rectangle 
contained by the means, the four straight lines are 
proportionals. 

Let the four straight lines AB, CD, E, F, be proportionals* 
viz. as AB to CD, so E to F ; the rectangle contained by AB, 
F is equal to the rectangle contained by CD, E. 

From the points A, C draw* AG, CH at right angles to 
AB, CD ; and make AG equal to F, and CH equal to E, and 
complete theparallellograms BG, DH : Because, as AB to CD, 
so is E to F ; and that E is equal to CH, and F to AG; AB 
is'' to CD as CH to AG. Therefore the sides of the paralle- 
lograms BG, DH about the equal angles are reciprocally pro- 
portional ; but parallelograms which have their sides about 
equal angles reciprocally proportional, arc equal to one ano.. 
ther*^ ; therefore the parallelogram BG is equal to the paralle- 
logram DH : And the paral- ^ 
lelogram BG is contained by ^' 
the straight lines AB, F; be- 
cause AG is equal to F ; and 
the parallelogram DH is con- 
tained by CD and E; because 
CH is equal to E ; There- 
fore the rectangle contained 
by the straight lines AB, F 
is equal to that which is 
contained by CD and E. 

And if the rectangle contained by the straight lines AB, t 
be equal to that which is contained by CD, E; these four lines 
are proportional, viz. AB is to CD, as E to F 

The same construction being made, because the rectang 
contained by the straight lines AB, F is equal to that which 
contained by CD, E, and that the rectangle BG is contained ' 
AB, F, because AG is equal to F ; and the rectangle DH by 
CD, E, because CH is equal to E ; therefore the parallelo- 
gram BiG is equal to the parallelogram DH j and they are 

equi- 




I 



OF EUCLID. 



165 



equiangular : But the sides about the equal angles of equal Boot VI. 
paralieloff rams-are reciprocally proportional'^ ; Wherefore, as ^"^T'^T^ 
AB to CD, so is CH to AG ; and CH is equal to E, and AG ' * 
to F: as therefore AB is to CD, so E to F. Wherefore, if 
four, &c. Q^E. D. 



PROP. XVII. THEOR. 



B- 



J.F three straight lines be proportionals, therectan- 
gle contained by the extremes is equal to the square 
of the mean : and if the rectangle contained by the 
extremes be equal to the square of the mean, the 
three straight lines are proportionals. 

Let the three straight lines A, B, C be proportionals, viz. 
as A to B, so B to C J the rectangle contained by A, C is equal 
to the square of B. 

Take D equal to B j and because as A to B, so B to C, and 
that B is equal to D • A is * to B, as D to C : But, if four 
straight lines be pro- 
portionals, the rectan- \ • 

gle contained by the 
extremes is equal to 
that which is contained 
by the means'': There- 
fore the rectangle con- 
tained by A, C is equal 
to that contained by 
B, D : But the rect- 
angle contained by B, 
D is the square of B ; because B is equal to D : Therefore 
the rectangle contained by A, C is equal to the square of B. 

And if the rectangle contained by A, C be equal to the 
square of B ; A is to B, as B to C. 

The same construction being made, because the rectangle 
contained by A, C is equal to the square of B, and the square 
of B is equal to the rectangle contained by B, D, because B 
is equal to D j therefore the rectangle contained by A, C is 
equal to that contained by B, D ; but if the rectangle con- 
tained by the extremes be equal to that contained by the means, 
the four straight lines are proportionals^ : Therefore A is to 

M 3 B, as 




7. #. 



* 16. 6. 



i66 O F E U C L I D. 

Book VI, B, as D to C i but B is equal to D ; wherefore, as A to B, 
'"^'"'''^^ so B to C i Therefore, if three straight lines, Sec. Q. E. D. 



SeeN. 



PROP. XVIII. PROB. 

U PON a given straight line to describe a rectili- 
neal figure similar, and similarly situated to a given 



rectilineal figure 



a" 



Let AB be the given straight line, and CDEF the given 
rectilineal figure of four sides; it is required upon the given 
straight line AB to describe a rectilineal figure similar, ^nd 
similarly situated to CDEF. 

Join DF, and at the points A, B in the straight line AB, 

■> :.5. 1. malce^ the angle BAG equal to the angle! at C, and the angle 

ABG equal to the angle CDF ; therefore the remaining angle 

■•'*" • CFD is equal to the remaining angle AGB^ Wherefore the 

triangle FCD is 



equiangular, to the jr 

l\ 




triangle GAB: (^ 
Again, at the points 
G, B in the straight 
line GB, make * tiie 
angle BGH equal to 
the angle DFE, and 
the angle G B H, A 
equal to F D E ; 
therefore the remaining angle FED Is equal to the remaining 
angle GHB, and the triangle FDE equiangular to the triangle 
GBH : Then, because the angle AGB is equal to the angle 
CFD, and BGH to DFE, the whole angle AGH is equal 
to the whole CFE: For the same reason, the angle ABH is 
equal to the angle CDE ; also the angle at A is equal to the 
angle at C, and the angle GHB to FED : Therefore the rec- 
tihneal figure ABHG is equiangular to CDEF : But likewise 
these figures have their sides about the equal angles propor- 
tionals ; because the triangles GAB, FCD being equiangular, 

.. ■;. BA is <= to AG, as DC to CF ; and because AG is to GB, as 
CF to FD ; and as GB to GH, so, b}' reason of the equiangu- 

■^2. :j. lar triangles BGH, UFE, is FD to FE; therefore, ex aequali'', 
AG is to GH, as CF to FE : In the same manner it may be 
proved that AB is to BH, as CD to DE : And GH is to 

HB, as 



OF EUCLID. 167 

HB, as FE to ED=. Wherefore, because the rectilineal figures Book vi. 
ABHG, CDEF are equiangular, and have theirsidesabout the ^•-^•'''*^ 
equal angles proportionals, they are similar to one anothor*. « i_ jigf. g. 

Next, let it be required to describe upon a given straight 
line AB, a rectilineal figure similar, and similarly situated to 
the rectilineal figure CDKEF. 

Join DE, and upon the given straight line AB describe the 
rectilineal figure ABHG similar, and similarly situated to the 
quadrilateral figure CDEF, by the former case j and at the 
points B, H, in the straight line BH, make the angle HBL 
etjual to the angle EDK, and the angle BHLequal to the angle 
DEK ; therefore the remaining angle at K is equal to the re- 
mainin;^ angle at L : And because the figures ABHG, CDEF 
are similar, the angle GHB is equal to the angle FED, and 
BHL is equal to DEK ; wherefore the w^hole angle GHL is 
equal to the whole angle FEK: For the same reason the angle 
ABL is equal to the angle CDK : Therefore the five -sided 
fio'ures AGHLB, CFEKD are equiangular ; and because the 
figures AGHB, CFED are similar, GH is to HB, as FE to 
ED ; and as HB to HL, so is ED to EK= ; therefore, ex sequa- 
li-, GH is to HL, as FE to EK : For the same reason, AB is ' *■ «• 
to BL as CD to DK : And BL is to LH, as-^ DK to KE, be- « 22. 5. 
causethe triangles BLH, DKF are equiangular : Therefore, 
because the five-sided figures AGHLB, CFEKD are equian- 
gular, and have their sides about the equal angles proportion- 
als, they are similar to one another ; and in the same manner 
a rectilineal figure of six or more sides may be described upon 
a given straight line similar to one given, and so on. Which 
was to be done. 



PROP. XIX. THEOR. 

Similar triangles are to one another in the du- 
plicate ratio of their homologous sides. 

Let ABC, DEF be similar triangles, having the angle B . 
equal to the angle E, and let AB be to BC, as DE to EF, so 
that the side BC is homologous to EF^ : the triangle ABC * ^2. def. 
has-fio the triangle DEF, the duplicate ratio of that which BC 
has to EF. 

Take BG a third proportional to BC, EF*': so that BC is*" 11. 6. 
to EF, as EF to BG, and join GA : Then, because as.AB 
to BC, so DE to EF j alternately^ AB is to DE, as BC to ' ic 5. 
M4 EF; 





i68 THEELEMENTS 

^K VI. EF : But as BC to EF, so is EF to BG ; thereforeJ as AB 

o II ^^ to DE, so is EF to BG : Wherefore the sides of the triangles 
ABG, DEF, which are about the equal angles, are recipro- 
cally proportional : But triangles which have the sides about 
two equal angles re- 
ciprocally proportio- j^ - 
nal are equal to one 

* 15. 6. another'^ : I'herefore 
the triangle ABG is 
equal to the triangle 
DEF: And because 
as BC is to EF, so 
EF to BG ; anh that 
if three straight lines « 

'10. def.5. be proportionals, the first is said *' to have to th6 third the du- 
plicate ratio of that which it has to the second ; BC therefore 
has toBG the duplicate ratio of that which BC has to EF : 

e I.e. ' But as BC to BG, so iss the triangle ABC to the triangle 
ABG. Therefore the triangle ABC has to the triangle ABG 
the duplicate ratio of that which BC has to EF : But the tri- 
angle ABG is equal to the triangle DEF : Wherefore also the 
triangle ABC has to the triangle DEF the duplicate ratio of 
that which BC has to EF. Tiierefore similar triangles, &c. 
Q, E. D. . 

Cor. From this it is manifest, that if three straight lines 
be proportionals, as the first is to the third, so is any triangle 
upon the first to a similar, and 'similarly described triangle 
upon the second. 

PROP. XX. THEOR. 

oLMILAR polygons may be divided into thesame 
number of similar triangles, having the same latio to 
one another that the polygons have; and the poly- 
gons have to one another the duplicate ratio of that 
which their liomoloo-ous sides have. 

Let ABCDE, FGHKL be similar polygons, and let AB be 
the honiologous side to FG: The polygons ABCDE, 
FGHKL may be divided into the same number of similar tri- 
angles, whereof each to each has the same ratio which the po- 
lygons have ; and the polygon ABCDE has to the polygon 
FGHKL the duplicate ratio of that which the side AB has to 
the side FG. 

Join BE, EC;GL, LH : And because the polygon ABCDE 

is 



OF EUCLID. 169 

is similar to thepolygon FGHKL, the angle BAE is equal to the ^'^• 
angle GFL*, and BA is to AE, as GF to FL=^: Wherefore, , ^ ^^j ^^ 
because the triangles ABE, FGL have an angle in one equal 
to an an^^le in the other, and their sides about these equal angles 
proportionals, the triangle ABE is equiangular'', and there- " 6. 6. 
fore similar to the triangle FGL<^ ; wherefore the angle ABE "i-S. ^ 
is equal to the angle FGL : And, because the polygons are simi- 
lar, the whole angle ABC is equalHo the wliole angle FGH ; 
therefore the remaining angle EBC is equal to the remaining 
angle LGii : And because the triangles ABE, FGL are similar, 
EB is to BA, as LG to GF»; and also, because the polygons 
are similar, AB is to BC, as EG to GH^ therefore, ex aequali'^, * 22. 5. 
EB is to BC, as LG to GH ; that is, the sides about the 
equal aneles EBC, LGH are proportionals; therefore'^ the 
tnano-le EBC is equiangular to the triangle LGH, and similar 
to its For 

the same rea- ^ 

son, the tri- ^,x^'^^s. M ^ 

angle ECD ^^ ^^^ X, r >> 

likewise is '^;^~ y ^ ^ 

similar to V^\ / TJw /" 

the triangle 
LHK: there- 
fore the si- 
milar poly- 
gons ABCDE, FGHKL are divided into the same number of 
similar triangles. 

Also these triangles have, each to each, the same ratio which 
the polygons have to one another, the antecedents being ABE, 
EBC, ECD, and the consequents FGL, LGH, LHK : And 
the polygon ABCDE has to the polygon FGHKL the dupli- 
cate ratio of that which the side AB has to the homologous 
side FG. 

Because the triangle ABE is similar to the triangle FGL, 
ABE has to FGL, the duplicate ratio<^ of that which the side « 19. o 
BE has to the side GL : For the same reason, the triangle BEC 
has to GLH the duplicate ratio of that which BE has to GL : 
Therefore, as the triangle ABE to the triangle FGL, so*^ is the^ n. ,. 
triangle BEC to the triangle GLH. Again, because, the tri- 
angle EBC is similar to the triangle LGH, EBC has to LGH, 
the duplicate ratio of that which the side EC has to the side 
LH : For the same reason, the triangle ECD has to the triangle 

LHK, 





lyo THE ELEMENTS 

Book VI. LHK, the duplicate ratio of that which EC has to LH : As 
>^r*^ therefore the triangle EBC to the triangle LGH, so is*" the 

'11. 5. . , ^ ^ 

triangle 

ECD to the A 

triangle ^^-'•^^\ M 

LHK: But 

it has been 
proved, that 
the triangle 
EBC is like- 
wise to the 
triangle 

LGH, as the triangle ABE to the triangle FGL. Therefore, 
as the triangle ABE to the triangle FGL, so is triangle EBC 
to triangle LGH, and triangle PXD to triangle LHK : And 
therefore, as one of the antecedents to one of the consequents, 

s 12. 5. so are all the antecedents to all the consequents s. Wherefore, 
as the triangle ABE to the triangle FGL, so is the polygon 
ABCDE to the polygon FGHKL : But the triangle ABE has 
to the triangle FGL, the duplicate ratio of that which the side 
AB has to the homologous side FG. Therefore also the poly- 
gon ABCDE has to the polygon FGHKL the duplicate ratio 
of that which AB has to the homologous side EG. Where- 
fore similar polygons, &c. Q^ E. D. 

Cor. I. in like manner, it may be proved, that similar four 
sided figures, or of any number of sides, are one to another in 
the duplicate ratio of their homologous sides, and it has already 
been proved in triangles. Therctore, universally, similar rec- 
tilineal figures are to one another in the duplicate ratio of their 
homologous sides. 

Cor. 2". And if to AB, FG, two of the homologous sides, 

h lO.dcf.5. a third proportional M be taken, AB has** to M the duplicate 
ratio of that which AB has to FG : but the four-sided figure or 
polygon upon AB, has to the four-sided figure or polygon upon 
FG likewise the duplicate ratio of that which AJ5 has to FG : 
Therefore, as AB is to M, so is the figure upon AB to the 

*Cor.j9.6. %'J''s upon FG, which was also proved in triangles'. 
Therefore, universally, it is manifest, that if three straight 
lines be proportionals, as the third is to the third, so is any rec- 
tilineal figure uptfn the first, tb a similar and similarly de- 
scribedrectilineal figure upon the second. 





OF EUCLlb. 



PROP, XXI. THEOR. 

X\ E c T I L i^N ZA L figuics wliich are similar to the same 
rectilineal figure, are also similar to one another. 

Let each of the rectilineal figures A, B be similar to the rec- 
tilineal figure C : The figure A is similar to the figure B. 

Because A is similar to C, they are equiangular, and also 
have their sides about the equal angles proportionals'. Again, » i. d*Li. 
because B is similar to 
C, they are equian- 
gular, and have their 
sides about the equal 
angles proportionals*. 
Therefore the figures 
A, B are each of them 

equiangular to C, and have the sides about the equal angles of 
each of them and of B proportionals. Wherefore the rectili- 
neal figures A and C are equiangular'', and have their sides " i. Ax. i. 
about the equal angles proportionals'". Therefore A is similar* "^ J'- J- 
to B. Q, E. D. ^ 

PROP. XXII. THEOR. 

J-F four straight lines be proportionals, the similar 
rectilineal figures similarly described upon them 
shall also be proportionals ; and it' the similar rec- 
tilineal figures similarly described upon four straight 
lines be proportionals, those straight lines shall be 
proportionals. 

Let the four straight lines AB, CD, EF, GH be propor- 
tionals, viz. AB to CD, as EF to GH, and upon AB, CD let 
the similar rectilineal figures KAB, LCD be similarly de- 
scribed ; and upon EF, GH the similar rectilineal figures 
VIF, NH, in like manner : The rectilineal figure KAB is to 
XD, as MF to NH. 

To AB, CD take a third proportional* X; and to EF, GH » ii. «. 
I third proportional O : And because AB, is to CD, as EF to 
3H, and that CD is'' to X, as GH to O; wherefore, ex Ml. 5. 

qiuliS as AB to X, so EF to O : But as AB to X, so is the c oo. 5. 

rectiliaeal 




172 THE ELEMENTS 

£ooK VI. rectilineal KAB tothfe rectilineal LCD, and as EF to O, so is 

^ the rectilineal MF to the rectilineal NH : Therefore, as 

KAB to LCD, so" is MF to N H. 

And if the reailineal KAB be to LCD, as MF to NH'; 

the straight line AB is to CD, as EF to GH. 
«12. 6. Malce'= as AB to CD, so EF to PR, and upon PR describe' 

the rectilineal figure SR similar and similarly situated to either 



X 



»"18. 6. 







E F <i H P K 



of the figures MF, NH : Then, because as AB to CD, so is 
EF to PR, and that upon AB, CD are described the similar and 
similarly situated rectilineals KAB, LCD, and upon EF, PR, 
in like manner, the similar rectilineals MF, SR; KAB is to 
LCD, as MF to SR; but by the hypothesis KAB is to LCD, 
as MF to NH ; and therefore the rectilineal MF having the 
t. 9, 5. same ratio to each of the two NH, SR, these are equal g to one 
another : They are also similar, and similarly situated ; there- 
fore GH is equal to PR : And because as AB to CD, so is EP 
to PR, and that PR is equal to GH ; AB is to CD, as EF tc 
GH. If therefore four straight lines, &c. Q. E. D. 



PROP. XXin. THEOR. 

SeeN. liQuiANGyLAR parallelogTams have to one ano- 
ther the ratio which is compounded of the ratios oS 
their sides. j 

Let AC, CF be equiangular parallelograms, having the angll 
BCD equal to the angle ECG: The ratio of the parallelogi am 
AC to the parallelogram CF,is the same with the ratio which 
is compounded of the ratios of their sides. 

L^ 
I 




OF EUCLID. 173 

• Let BC, CG be placed in a straight line; therefore DC and Book vi, 
CE are also in a straight line*; and complete the parallelogram ^^^ 
DG ; and talcing any straight line K, make*' as BC to CG, " 12! 6. 
so K to L ; and as DC to CE, so make^ L to M : Therefore, 
the ratios of K to L ; and L to M, are the same with the ratios 
of the sides, viz, of BC to CG, and DC to CE. But the ratio 
of K to M is that which is said to he compounded '^ of the « A. dcf. j. 
ratios of K to L, and L to M : Wherefore also K has to M 
the ratio compounded of the ratios 
of the sides : And because as BC 
to CG, so is the parallelogram AC 

to the parallelogram CH' ; but \ \ \ • «ii. c. 

as BC to CG, so is K to L ; there- 

fore K is ^ to L, as the parallel- 1 \ \ «ii.5. 

ogram AC to the parallelogram 
CH : Again, because as DC to 
CE, so is the parallelogram CH 
to the parallelogram CF ; but as 
DC to CE, so is L to M ; where- 
fore L is f to M, as the paral- 
lelogram CH to the parallelogram CF : Therefore since it has 
been proved, that as K to L, so is the parallelogram AC to the 
parallelogram CH; and as L to M, so the parallelogram CH 
to tr.e parallelogram CF ; ex aequali*^, K is to M, as the pa- f 33. 5, 
rallelogram AC to the parallelogram CF : But K has to M the 
rat'io which is compounded of the ratios of the sides; therefore 
also the parallelogram AC has to the parallelogram CF the 
ratio which is compounded of the ratios of the sides. Where- ' 
fore ecjuiangular parallelograms, &c. Q. E. D. 



PROP. XXIV. THEOR. 



i HE parallelograms about the diameter of any sceN. 
parallelogram, are similar to the whole, and tp one 
another. 

Let ABCD be a parallelogram, of which the diameter is 
AC ; and EG, HK the parallelograms about the diameter : 
The parallelograms EG, HK are similar both to the whole 
parallelogram ABCD, and to one another. 

Because DC, GF are parallels, the angle ADC is equal* to » 2?. 1. 
•he angle AGF : For the same reason, because BC, EF are pa- 
rallels, 



174 



*. 6. 



«1. clef. 6. 



n-.e. 



SeeN. 




"Cor. 45.1 



129, 



129, 1. 
1, 
« 13. 6. 
■» 18. 1. 



' 2. Cor. 
2(). 6. 



THE ELEMENTS 

rallels, the angle ABC is equal to the angle AEF : And eacK 
of the angles BCD, EFG is equal to the opposite angle DAB*", 
and therefore are equal to one another, wherefore the paral- 
lelograms ABCD, AEFG are equiangular : And because the 
angle ABC is equal to the angle AEF, and the angle BAG 
common to the two triangles BAC, EAP\ they are equiangu- 
lar to one another ; therefore*^ as AB 
toBC, sois AEtoEF: And because 
the opposite sides of parallelograms 
are equal to one another^, AB'' is 
to AD, as AE to AG ; and DC to 
CB as GF to FE ; and also CD to 
DA, as FG to GA ; Therefore the 
sides of the parallelograms ABCD, 
AEFG about the equal angles are 
proportionals ;' and they are there- 
fore similar to one another^ : For the same reason the paral- 
lelogram ABCD is similar to the parallelogram FHCK. 
Wherefore each of the parallelograms GE, KH is similar to 
DB : But reftilineal figures which are similar to the same 
rectilineal figure are also similar to one another*^ ; therefore 
the parallelogram GE is similar to KH. Wherefore the pa. 
rallelograms, &c. Q^ E. D. 

'PROP. XXV. PROB. 

1 O describe a rectilineal figure which sliall be si- 
milar to one, and equal to another given rectilineal 
figure. 

Let ABC be the given rectilineal figure, to which the figure 
to be described is required to be similar, and D that to which 
it must be equal. It is required to describe a rectilineal figure 
similar to ABC, and equal to D. 

Upon the straight line BC describe'' the parallelogram BE 
equal to the figure ABC; also upon CE describe '^ the paral- 
lelogram CM equal to D, and having the angle FCE equal 
to the angle CBL: Therefore BC and CF are in a straight 
line^, as also LE and EM : Between BC and CF find"^ a mean 
proportional GH, and upon GH describe'' the redilineal 
figure KGH similar and similarly situated to the figure ABC: 
And because BC is to GH as GH to CF, and if three straight 
lines be proportionals, as the first is to the third, so is '^ the 

%ure 



OF EUCLID. 



figure upon the first to the similar and similarly described fi- ^°ok vi. 
gure upon the second ; therefore as BC to CF, so is the rec- ^^^ 
tilineal figure ABC to KGH : But as BC to CF, so is*^ the pa- *" i. 6. 
rallelogram BE to the parallelogram ^F : Therefore as the 
rectilineal figure ABC is to KGH, so is the parallelogram BE 
to the parallelogram EFs ; and the rectilineal figure ABC is« ji. 5. 

A 




equal to the parallelogram BE ; therefore the rectilineal figure 
KGH is equal'' to the parallelogram EF : But EF is equal' 
to the figure D ; wherefore also KGH is equal to D ; and it 
is similar to ABC Therefore the rectilineal figure KGH 
has been described similar to the figure ABC, and equal to 
D. Which was to be done. 



u, 5, 



PROP. XXVI. THEOR. 

IF twosimilar parallelograms have a common an- 
gle, and be ;^imilarly situated ; they are about the 
same diameter. 

Let the parallelograms ABCD, AEFG be similar and simi- 
larly situated, and have the angle DAB common. ABCD and 
AEFG are about the same diameter. 

For, if not, let, if possible, the 
parallelogram BD have its dia- 
j meter AHC in a different straight 
! line from AF, the diameter of the 
I parallelogram, EG, and let GF 
meet AHC in H; and throuc^h 
jH draw HK parallel to AD or 
{ BC : Therefore the parallelograms 
I ABCD, /\KHG being about the 
same diameter, they are similar to 
©ne another^ : Wherefore as DA to AB, so is*^ GA to AK : 

2 But 




• 24. «. 

» l.-def. C. 



176 THE ELEMENTS 

i;ooK VI. But bqcause ABCD and AEFG arc similar parallelograms, 
^"^^^^^ as DA is to AB, so is GA to AE j therefore'^ as GA to AE, 
so GA to AK J wherefore GA has the same ratio to each of 
* 9. 5. the straight lines AE, AK ; and consequently AK is equal^ to 
AE, the less to the greater, which is impossible : Therefore 
ABCD and AKHG are not about the same diameter ; where- , 
fore ABCD and AEFG must be about the same diameter. 
Therefore if two similar, &c. Q^ E. D. 

' To understand the three following propositions moreea- 

* sily, it is to be observed, 

* 1. That a parallelogram is said to be applied to a straight 
' line, when it is described upon it as one of its sides. Ex. gr. 
' the parallelogram AC is said to be applied to the straight 

* line AB. 

' 2. But a parallelogram AE Is said to be applied to a straight 

* line AB, deficient by a parallelogram j when AD the base of 

* AE is less than AB, and there- 
' * fore AE is less than the paral- 

' lelogram AC described upon 

* AB in the same angle, and 
' between the same parallels, by 
' the parallelogram DC; and 
' DC is therefore called the 
'defect of AE. 

* 3. And a parallelogram AG is said to be applied to a 
' straight line AB, exceeding by a parallelogram, when" AF 
' the base of AG is greater than AB, and therefore AG ex- 
' ceeds AC the parallelogram described upon AB in the same 
' angle, and between the same parallels, by the parallelogram 

* BG.' 



5CC N, 





E ( 


c 


G 


" 






A. 


i) J 


s 


V 



PROP. XXVII. THEOR. 

v^F all parallelograms applied to the same 
straight line, and deficient by parallelograms, simi- 
lar and similarly situated to thatwliichis described 
upon the half of the line ; that which is applied to 
the half, and is similar to its defect, is the greatest. 

Let AB be a straight line divided into two equal parts inC, 
and let the parallelogram AD be applied to the half AC, 
which is therefore deficient from the parallelogram upon the 
whole line AB by the parallelogram CE upon the other half 

CB: 



OF EUCLID. 

CB : Of all the parallelograms applied to any other parts of 
AB, and deficient by parallelograms that are similar, and simi- 
larly situated to CE, AD is the greatest. 

Let AF be any parellelogram applied to AK, any other 
part of AB than the half, so as to be deficient from the parallel- 
ogram upon the whole line AB by the parallelogram KH si- 
mitar and similarly situated toCE : AD is greater than AF, 

First, let AK the base of AF, be greater than AC the half of 
AB ; and because CE is similar to the 
parallelogram KH,they are about the 
same diameter* : Draw their diameter 
DB, and complete the scheme: Be- 
cause the parallelogram CF is equal*" 
to FE, and KH to both, therefore the 
whole CH is equal to the whole KE : 
But CH is equa^ to CG, because the 
base AC is equal to the base CB; 
therefore CG is equal to KE : To 
each of these add CF i then the whole 

AF is equal to the gnomon CHL ; Therefore CE, or the 
parallelogram AD, is greater than the parallelogram AF. 

Next, let AK, the base of AF, be 
less than AC, and, the same construc- 
tion being made, the parallelogram 
DH is equal to DGs for HM is 
equal to MG'', because BC is equal 
to CA ; wherefore DH is greater 
thanLG: But DH is equalno DK j 
therefore DK is greater than LG : 
To each of these add AL ; then the 
■whole AD is greater than the whole 
AF. Therefore, of all parallelograms 
applied, &c. Q. E. D. 



J77 



Book VI. 




^36. I. 



KB 




*^. 1. 



N 



178 



Book VI. 



Sec N. 



» 10. 1. 
18. 6. 



•< 21. 6. 



THE ELEMENTS 



PROP. XXVIII. PROB. 



J. O a given straight line to apply a parallelogram 
equal to a given rectilineal figure, and deficient by 
a parallelogram similar to a given parellelogram : 
But the given rectilineal figure to which the paral- 
lelogram to be applied is to be equal, must not be 
greater than the parallelogram applied to half of the 
given line, having its defect similar to the defect of 
that which is to be applied ; that is, to the given 
parallelogram. 

Let AB be the'given straight line, and C the given rectilineal 
figure, to which the parallelogram to be applied is required 
to be equal, which figure must not be greater than the paral- 
lelogram applied to the half of the line having its defect from 
that upon the whole line similar to the defect of that which 
is to be applied ; and let D be the parallelogram to which this 
defect is required to be similar. It is required to apply a pa- 
rallelogram to the straight 

line AB, which shalfbe H G_ OF 

equal to the figure C, and 
be deficient from the paral- 
lelogram upon the wioJe 
line by a parallelogram si- 
milar to D. 

Divide AB into two equal 
f arts=» in the point E, and 
upon EB describe the pa- 
rallelogram EBFG similar'' 
and similarly situated to D, 
and complete the parallelo- 
gram AG, which must 
cither be equal to C, or greater than it, by the determination : 
And if AG be squal to C, then what was required is already 
dond : For, upon the straight line AB, the parallelogram AG 
is applied equal to the figure C, and deficient by the parallelo- 
gram EF similar to D : But, if AG be not equal to C, it is 
greater than it ; and EF is equal to AG ; therefore EF also 
is greater than C. Make c the parallelogram KLMN equal to" 
the excess of EF above C, and similar and similarly situated 
to D i but D is similar to EF, therefore* also KM is similar 

I to 




OF EUCLID. 179 

to EF : Let KL be the homologous side to EG, and LM to Book VL 
GF : and because EF is equal to C and KM together, EF ^^'^''^^ 
is greater than KM ; therefore the straight line EG is greater 
than KL, and GF than LM : Make GX equal to LK, and 
GO equal to LM, and complete the parallelogram XGOP : 
Therefore XO is equal and similar to KM ; but KM is simi- 
lar toEF ; wherefore also XO is similar to EF, and therefore 
XO and EF are about the same diameter* : LetGPB be their * -6. o. 
diameter, and complete the scheme : Then because EF is 
equal to C and KM together, and XO a part of the one is 
equal to KAI a part of the other, the remainder, viz. the gno- 
mon ERO, is equal to the remainder C : and because OR is 
equal ^ to XS, by adding SR to each, the whole CB is equal f 34 2. 
to the whole XB : But XB is equals to TE, because the base g 3^ ^ 
AE is equal to the base EB j wherefore also TE is equal to 
OB; Add XS to each, then the whole TS is equal to the 
whole, viz, to the gnomon ERO : But it has been proved 
that the gnomon ERO is equal to C, and therefore also TS is 
equal to C. Wherefore the parallelogram TS, equal to the 
given rectilineal figure C, is applied to the given straight line 
AB deficient by the parallelogram SR, similar to the given one 
D, because SR is similar to EF**. Which was to be done. " 24. «, 



PROP. XXIX. PROB. 



JL O a given straight line to apply a parallelogram seeN. 
equal to a given rectilineal figure, exceeding by a 
parallelogram similar to another given. 

Let AB be the given straight line, and C the given rectili- 
neal figure to which the«parallelogram to be applied is required 
to be equal, and D the parallelogram to which the excess of 
the one to be applied above that upon the given line is required 
to be similar. It is required to apply a parallelogram to the 
given straight line AB which shall be equal to the figure C, 
exceeding by a parallelogram similar to D. 

Divide AB into two equal parts in the point E, and upon » 13. e, 
EB describe ' the parallelogram EL similar, and similarly situ- 

N 2 ated 



i8o THE ELEMENTS 

Book VI. ated to-D : And make ^ the parallejogram GH equal t6 EL and 
C together, and similar, and similarly situated to D ; where- 
fore CH is similar to EL*^ : Let KH be the side homologous 
/ to FL, and KG to FE : And because the parallelogram GH 

is greater than EL, therefore the side KH is greater than FL, 
and KG than FE : Produce FL and FE, and make FLM equal 
to KH,and FEN toKG,and complete the parallelogram MN. 
MN is therefore equal 
and similar to GH j 

. ' but GH is similar to 

EL ; wherefore MN 
is similar to EL, and 
consequently EL and 
MN are about the 

"26.6. same diameter^: Draw 
their diameter FX, 
and complete the 
scheme. Therefore, 
.since GH is equal to 
EL and C together, 
and that GH is equal 
toMN J MN is equal 
to EL and C : Take away the common part EL ; then the 
remainder, via. the gnomon NOL, is equal to C. And be- 
cause AE is equal to EB, the parallelogram AN is equal= to 
the parallelogram NB, that is, to BM^. Add NO to each ; 
therefore the whole, viz. the parallelogram AX, is equal to the 
gnomon, NOL^ But the gnomon NOL is equal to C ; there- 
fore also AX is equal to C. Wherefore to the straight line 
AB there is applied the parallelogram AX equal to the given 
rectilineal C, exceeding by the parallelogram PO, which i& 

• 24. 6. similar to D, because PO is similar to ELs. Which was t* 
be done. 



• 36. 1. 
'«. 1. 




PROP. XXX. PROB. 



To 

ratio. 



cut a given straight line in extreme and mean 



Let AB be the given straight linej it is required to cut it 
in extreme and mean ratio. 

1 Upon 



B 



OF EUCLID. 

Upon AB describe » the square BC, and to AC apply the 
parallelogram CD equal to BC, exceeding by the %ure AD 
similar to BC'': But BC is a square, 
therefore also AD is a square j and be- 
cause BC is equal to CD, by taking the 
common part CE from each, the re- 
mainder BF is equal to the remainder J^ 
AD : And these figures are equiangular, 
therefore their sides about the equal an- 
gles are reciprocally proportional c : 
Wherefore, as ¥E to ED, so AE to EB : 
But FE is equal to AC\ that is, to AB ; 
and ED is equal to AE : Therefore as 
BA to AE, so is AE to EB : But AB ^ 
is greater than AE ; wherefore AE is 
greater than EB= : Therefore the straight line AB is cut in 
extreme and mean ratio in E'. Which was to be done. 
Otherwise, 

Let AB be the given straight line ; it is required to cut it 
in extreme and mean ratio. 

Divide AB in the point C, so that the rectangle contained 
by AB, BC be equal to the square of AC s : 

Then, because the rectangle AB, BC is -y: p p 

equal to the square of AC, as BA to AC, 

so is AC to CB ^ : Therefore AB is cut in extreme and mean 

ratio in C*^. Which was to be done. 



iSi 

Book VI. 





E ■ 








' 



' 14. 6, 

'^ .34. 1. 



« H. 5. 
'^ J def. 6. 



t tl. Ot 



» n. 6. 



PROP. XXXL THEOR. 



IN right angled triangles, the rectilineal figure de- See^u 
jcribed upon the side opposite to the right angle, is 
?qual to the similar, and similarly described figures 
ipon the sides containing the right angle. 

Let ABC be a right angled triangle, having the right angle 
5AC : The rectilineal figure described upon BC is equal to 
le similar, and similarly described figures upon BA, AC' 
Draw the perpendicular AD ; therefore, because in the right 
igled triangle ABC, AD is drawn from the right angle at A 
erpendicular to the base BC, the triangles. ABD, ADC are 
limilar to the whole triangle ABC, and to one another^ and i g. g. 

N 3 because 



182 



' 2 Cor. 
00. 6. 



THE ELEMENTS 

because the triangle ABC is similar to ADB, as CB to BA, s© 
is BA to BD '' ; and because these three straight lines are pro- 
portionals, as the first to the third, so is the figure upon the 
first to the similar, and similarly described figure upon the 
second^ : Therefore as CB to 
BD, so 'is the figure 
CB to the similar and 
larly described figure 



upon 
simi- 
upon 
* B. 5. BA': And inversely'', as DB 
to BC, so is the figure upon 
BA to that upon BC : For 
the same reason, as DC to 
CB, so is the figure upon CA 
to that upon CB. Wherefore 
as BD and DC t9gether to BC, so are the figures upon BA, 
e 2i. 5. AC to that upon BC " : But BD and DC together are equal 
f A. 5. to BC. Therefore the figure described on BC is equaK to the 
similar and similarly described figures on BA, AC. Where- 
fore, in right angled triangles, &c. Q^. E. D. 




PROP. XXXII. THEOR. 



SeeN. JLy two tHanglcs which have two sides of the one 
proportional to two sides of the other, be joined at 
one angle, so as to have their homologous sides pa- 
rallel to one another ; the remainins: sides shall be 
in a straight line. 

Let ABC, DCE, be two triangles which have the two sides: | 
BA, AC proportional to the two CD, DE, viz. BA to AC, as 
CD to DE ; and let AB be parallel to DC, and AC to DE,- 
BC and CE are in a straight line. 

Because AB is parallel to \ 
DC, and the straight line 
AC meets them, the alter- 
nate angles BAC, ACD 
* 29. 1, sre equal* ; for the same 
reason, the angle CDE is 
equal to the angle ACD ; 
wherefore also BAC is equal 
to CDE : And because 




OF EUCLID. 183 

the triangles ABC, DCE have one angle at A equal to one at ^poR vi. 
D, and the sides about these angles proportionals, viz. BA to ^''^^''^'^■ 
AC, as CD to DE, the triangle ABC is equiangular'' to DCE : " 6. 6. 
Therefore the angle ABC is equal to the angle DCE : And 
the angle BAC was proved to be equal to ACD : Therefore 
the whole angle ACE is equal to the two angles ABC, BAC ; 
add the common angle ACB, then the angles ACE, ACB are 
equal to the angles ABC, BAC, ACB : But ABC, BAC, 
ACB are equal to two right angles'^ ; therefore also the angles « 352. 1. 
ACE, ACB are equal to two right angles : And since at the 
point C, in the straight line AC, the two straight lines BC, 
CE, which are on the opposite sides of it, make the adjacent 
angles ACE, ACB equal to two right angles ; therefore"* BC " i-^- J- 
and CE are in a straight line. Wherefore, if two triangles, 
&c. Q. E. D. 



PROP. XXXIII. THEOR. 

J. N equal circles, angles, whether the centres' or seeN. 
circumferences, have the same ratio which the cir- ' 
cumferenccs on which they stand have to one ano- 
ther : So also have the sectors. 

Let ABC, DEF be equal circles ; and at their centres the 
angles BGC, EHF, and the angles BAC, EDF, at their cir- 
cumferences J as the circumference BC to the circumference 
EF, so is the angle BGC to the angle EHF, and the angle 
BAC to the angle EDF ; and also the seclor BGC to the 
sector EHF. 

Taiceany number of circumferences CK, KL, each equal to 
BC, and any number whatever FM, MN, each equal to EF : 
"And join GK, GL, HM, HN. Because the circumferences 
BC, CK, KL are all equal, the angles BGC, CGK, KGL 
are also all equal*: Therefore what multiple soever the circum^ * 27.3. 
ference BL is of the circumference BC, the same multiple is 
the angle BGL of the angle BGC : For the same reason, what- 
ever multiple the circumference EN is of the circumference 
EF, the same multiple is the angle EHN of the angle EHF ; 

N4 And 



i84 THE EI^EMENTS 

Book VI. And if the circumference BL be equal to the circumference 
.^"^iPg"**^ EN, the angle BGL is also equal'' to the angle EHN ; and 
if the circumference BL be greater than EN, likewise the angle 
BGL is greater than EHN ; and if less, less : There being then 
four magnitudes, the two circumferences BC, EF, and the 
two angles BGC, EHF ; of the circumference BC, and of the 
angle BGC, have been taken any equimultiples whatever, viz. 
the circumference BL, and the angle BGL ; and of the circum- 
ference EF, and of the angle EHF, any equimultiples what- 




ever, viz. the circumference EN, and the angle EHN : And 
it has been proved, that, if the circumference BL, be greater 
than EN, the angle BGL is greater than EHN j and if equal, 
equal; and if less, less: As therefore the circumference BC 

•• 5. def. 5. to the circumference EF, so ^ is the angle BGC to the angle 
EHF : But as the angle BGC is to the angle EHF, so is 

« 13. 5. *" the angle BAC to the angle EDF ; for each is double of 

<20. 3. cach^; Therefore, as thel circumference BC is to EF, so is 
the angle BGC to the angle EHF, and the angle BAC to the 
angle EDF. 

Also, as the circumference BC to EF, so is the sedlor BGC 
to the senior EHF. Join BC, CK, and in the circumferences 
BC,CKtake any points X, O, and join BX, XC, CO, OK: 
Then, because in the triangles GBC, GCK the two sides BG^ 
GC are equal to the two CG, GK, and that they contain 

• 4. 1. equal angles ; the base BC is equal = to the base CK, and the 
triangle GBC to the triangle GCK : And because the circum- 
ference BC is equal to the circumference CK, the remaining 
part of the whole circumference of the circle ABC, is equal to 
the remaining part of the whole circumference of the same 
circle : Wherefore the angle BXC is equal to the angle COK»; 

'n.^def.3. and the segment BXC is therefore similar to the segment COK*^; 

and 



OF EUCLID. 



iSs 



and they are upon equal straight lines BC, CK : But similar ^°°^ '^'• 
segments of circles upon equal straight lines, are equals to one ^H^^'^ 
another: Therefore the segment BXC is equal to the segment 
COK : And the triangle BGC is equal to the triangle CGKj 
therefore the whole, the sector BGC, is equal to the whole, the 
SecSlor CGK : For the same reason, the senior KGL is equal to 
each of the seftors BGC, CGK : In the same manner, the 
sedors EHF, FHM, MHN may be proved equal to one ano- 
ther : Therefore, what multiple soever the circumference BL 
is of the circumference BC, the same multiple is the sector BGL 
of the se£lor BGC : For the same reason, whatever multiple 
the circumference EN is of EF, the same multiple is the seftor 
EHN of the sedor EHF : And if the circumference BL be 




B-^: 



equal to EN, the sector BGL is equal to the sedor EHN ; and 
if the circumference BL be greater than EN, the sedtor 
BGL is greater than the sedor EHN; and if less, less: 
Since then, there are four magnitudes, the two circum- 
ferences BC, EF, and the two sedors BGC, EHF, and of 
the circumference BC, and sedor BGC, the circumference 
BL and sedor BGL are any equalmultiples whatever; and of 
.the circumference EF, and sedor EHF, the circumference 
EN, and sedor EHN, are any equimultiples whatever ; and 
that it has been proved, if the circumference BL be greater 
than EN, the sedor BGL is greater than the sedor EHN; 
'and if equal, equal ; and if less, less. Therefore,^ as the cir- 
cumference BC is to the circumference EF, so is the sedor 
BGC to the sedor EHF. Wherefore, in equal circles, kc, 
Q; E. D.- 



" 5. def. 5. 



i86 THE ELEMENTS 

Book VI. 

""^^ PROP. B. THEOR. 

See N, 1 F an angle of a triangle be bisected by a straight line, 
which likewise cuts the base ; the rectangle contained 
by the sides of the triangle is equal to the rectangle 
contained by the segments of the base, together with 
the square of the straight line bisecting the angle. 

Let ABC be a triangle, and let the angle BAC be bise£led 
by the straight line AD ; the reftangle BA, AC is equal to the 
rei3:angle BD, DC, together with the square of AD. 

' ^- *• Describe the circle* ACB about the triangle, and produce 

AD to the circumference in E, 
and join EC: Then because the 
angle BAD is equal to the angle 
CAE, and the angle ABD to the 

*2i.3. angle'' AEC, for they are in the 
samesegment; the triangles ABD, 
AEC, are equiangular to one ano- 
ther : Therefore as BA to AD, 

= 4.6. so is "^ EA to AC, and conse- 
quently the re£l:angle BA, AC is 

«• 16. 6. equaP to the redangle EA, AD, 

' 3. 2. that is= to the redtangle ED, DA, 

together with the square of AD ; But the re£langle ED, DA 

* 35. 3. is equal to the redtangle*^ BD, DC. Therefore the re£langle 
BA, AC is equal to the redangle BD, DC, together with the 
square of AD. Wherefore, if an angle, &cc. Q^ E. D. 

PROP. C. THEOR. 

seeN. J^p f^QYYi any angle of a triangle a straight line be 
drawn perpendicular to the base ; the rectangle con- 
tained by the sides of the triangle is ecjual to the 
rectangle contained by the perpendicular and the 
diameter of the circle described about the triano-lc. 




'O" 



Let ABC be a triangle, and AD the perpendicular from the 
angle A to the base BC j the re6langle B A, AC is equal to the 
rectangle contained by AD, and the diameter of the circle jde- 
scribed about the triangle. 

Describe 



OF EUCLID 



Describe* the circle ACB 
about the triangle, and draw its 
diameter AE, and join EC: Be- 
cause the right angle BDA is 
equal*" to the angle ECA in a se- 
micircle, and the angle ABD to 
the angle AEC in the same seg- 
ment"^ T the triangles ABD, AEC 
are equiano-ular : Therefore as 
«« BA to AD, so is EA to AC ; 
and consequently the retSbngle 
BA, AC is equal "^ to the red- 




16.6. 



angle EA, AD. If therefore from an angle, &c. Q. E. D. 



PROP. D. THEOR. 

X HE rectangle contained bv the dia2:onals of asceN, 
quadrilateral inscribed in a circle, is equal to both 
the rectangles contained by its opposite sides. 

Let ABCD be any quadrilateral inscribed in a circle, and 
join AC, BD j the rectangle contained by AC, BD is equal to 
the two rectangles contained by AB, CD, and by AD, BC*. 

Make the angle ABE equal to the angle DBC ; add to each 
of these the common angle EBD, then the angle ABD is equal 
to the angle EBC : And the angle BDA is equal* to the 
angle BCE, because they are in the same segment : therefore 
the triangle ABD is equiangular to 
the triangle BCE: Wherefore'', 
as BC is to CE, so is BD to DA i 
and consequently the rectangle BC, 
AD is equal*^ to the rectangle BD, 
CE : Again, because the angle 
ABE is equal to the angle DBC, 
,and the angle* BAE to the angle 
BDC, the triangle ABE is equi- 
angular to the triangle BCD : As 
therefore BA to AE, so is BD to 

DC; wherefore the rectangle BA, DC is equal to the rectangle 
BD, AE : But the rectangle BC, AD has been shewn equal 
to the rectangle BD, CE ; therefore the whole redtangl* AC, 
BD'' is equal to the rectangle AB, DC, together with the 
redangle AD, BC. Therefore the reaangle, &c. Q. E. D. 

♦This is a Lemma of CI. Ptolomseus, in page 9, of iys u-c^aKn mret^n. 




21. 3. 



4.6. 



M6. 6. 



d 1, O. 



Book XL 



t x88 ] 

THE 

EL E M E N T S 

OF 

EUCLID. 

BOOK. XI. 
DEFINITIONS. 

I. 

J\ SOLID is that which hath length, breadth, and thickness, 

11. 
That which bounds a solid is a superficies. 

III. 
A straight line is perpendicular, or at right angles to a plane, 
when it makes right angles with every straight line meeting 
it in that plane. 

IV. 
A plane is perpendicular to a plane, when the straight lines 
drawn in one of the planes perpendicularly to the common 
section of the two planes, are perpendicular to the other 
plane. 

V. 
The inclination of a straight line to a plane is the acute angle 
contained by that straight line, and another drawn from the 
point in which the first line meets the plane, to the point in 
which a perpendicular to the plane drawn from any point 
of the first line above the plane, meets the same plane. 

VI. 
The inclination of a plane to a plane is the acute angle con- 
tained by two straight lines drawn fr©m any the same point 
of their common section at right angles to it, one upon one 
plane, and the other upon the other. plane. 

VII. Two 



THE ELEMENTS OF EUCLID. 189 

VII. Book XI. 

Two planes are said to have the same, or alike inclination to ^'^^' 
one another, which two other planes have, when the said 
aneles of inclination are equal to one another. 
VIII. 
Parallel planes are such which do not meet one another though 
produced. 

IX. 
A solid angle is that which is made by the meeting of more g^ ^ 
than two plane angles, which are not in the same plane, in 
one point. 

X. 
* The tenth definition is omitted for reasons given in the notes.' See N. 

XL 
Similar solid figures are such as have all their solid angles See N. 

equal, each to each, and which are contained by the same 
- number of similar planes. 

XIL 
A pyramid is a solid figure contained by planes that are con- 
stituted betwixt one plane and one point above it in which 
they meet. 

XIII. 
A prism is a solid figure contained by plane figures, of which 
two that are opposite are equal, similar, and parallel to one 
another : and the others parallelograms, 
XIV. 
A sphere is a solid figure described by the revolution of a se- 
micircle about its diameter, which remains unmoved. 

XV. 
The axis of a sphere is the fixed straight line about which the 
semicircle revolves. 

XVI. 
The centre of a sphere is the same with that of the semicircle. 

XVII. 
The diameter of a sphere is any straight line which passes 
through the centre, and is teraainated both ways by the su- 
perficies of the sphere. 

XVIII. 
A cone is a solid figure described by the revolution of a right 
angled triangle about one of the sides containing the right 
angle, which side remains fixed. 
If the fixed side be equal to the other side containing the right 
angle, the cone is called a right angled cone ; if it be less 
than the other side, an obtuse angled, and if greater, an 
acute angled cone, 

XIX. The 




THE ELEMENTS 

XIX. 

The axis of a cone is the fixed straight line about which the 
triangle revolves. 

XX. 
The base of a cone is the circle described by that side contain- 
ing the fight angle, which revolves. 
XXI. 
A cylinder is a solid figure described by the revolution of a 
right angled parallelogram about one of its sides wrhich 
remains fixed. 

XXII. 
The axis of a cylinder is the fixed atraight line about which 
the parallelogram revolves. 

XXIII. 
The bases of a cylinder are the circles described by the two 
revolving opposite sides of the parallelogram. 
XXIV. 
Similar cones and cylinders are those which have their axes 
and the diameters of their bases proportionals. 
XXV. 
A cube is a solid figure contained by six equal squares, 

XXVI. 
A tetrahedron is a solid figure contained by four equal and 
equilateral triangles. 

XXVII. 
An octahedron is a solid figure contained by eight equal and 
equilateral triangles. 

XXVIII. 

A dodecahedron is a solid figure contained by twelve equal 

pentagons which are equilateral and equiangular. 

XXIX. 

An icosahedron is a solid figure contained by twenty equal 

and equilateral triangles. 

DEF. A. . 

A parallelopiped is a solid figure contained by six quadrilate- 
ral figures, whereof every opposite two are parallel. 




OF EUCLID. 191 

Book XI. 

PROP. I. THEOR. 

UnE part of a straight line cannot be in a plane, seex. 
and another part above it. 

If it be possible, let AB, part of the straight line ABC, be in 
the plane, and the part BC above it : And since the straight 
line AB is in the plane, it can be 
produced in that plane: Let it be 
propucedtoD: And let any plane 
pass through the straight line 
AD, and be turned about it un- 
til it pass through the point C ; 
and because the points B, Care in this plane, the straight line 
BC is in it^: Therefore there are two straight lines ABC, * ^dcf-l. 
ABD in the same plane that have a common segment AB, 
which is impossible''. Therefore, one part, &c. Q. E. D. i»Cor. 11.1 



PROP. n. THEOR. 

1 WO straight" lines ■which cut one another are in 
one plane, and three straight lines which meet one 
another arc in one plane. 

Let two straight lines AB, CD, cut one another in E ; AB» 
CD are one plane : and three straight lines EC, CB, BE^ 
which meet one another are in one plane. 

Let any plane pass through the straight 
line EB, and let the plane be turned 
about EB, produced, if necessary, until it 
pass through the point C : Then because 
the points E, C are in this plane, tthe 

straight line EC is in it*: For the same ^y\ ' '^^*'"'' '' 

reason, the straight line BC is in the 
same; and, by the hypothesis, EB is in 
it : Therefore the three straight lines EC, 
CB, BE are in one plane : But in the 
plane in which EC, EB are, in the same 
are" CD, AB: Therefore AB, CD, are j. ,;, 

in one plane. Wherefore two straight lines, &c. Q^ E. B. 





THE ELEMENTS 



PROP. III. THEOR. 

seeN. JLF two plaiics cut onc another, their common sec- 
tion is a straight Hne. 

Let two planes AB, 'BC, cut one another, and let the 
line DB betiieir common section: DB is 
a straight line : If it be not, from the 
point p to B, draw, in the plane AB, the 
straight line DlEB, and in the plane BC, 
the straight line DFB : Then two 
straight lines DEB, DFB have the same 
extremities, and therefore include a space 
*jOAx. 1. betwixt them; which is umpossibie^ : 
Therefore BD the common section of 
the planes AB, BC, cannot but be a 
straight line. Wherefore, if two planes, 
&c. Q, E. D. 



.B 



E 



DN 



A^ 



SeeN. 



« 15. 1. 
^ 4. J. 



« 26. I. 



PROP. IV. THEOR. 

IF a straight line stand at right angles to each of 
two straight lines in the point of their intersection, 
it shall also he at right angles to the plane which 
passes through them, that is, to the plane in which 
they are. ■ 

Let the straight line EF stand at right angles to each of the 
straight lines AB, CD in E, the point of their intersection : 
EF is also at right angles to the plane passingthrough AB, CD. 

Take the straight lines AE, EB, CE, ED,'all equal to one an- 
other; and through E draw, in the plane in which are AB,CD, 
any straight line GEH ; and join AD, CB ; then, from any 
point F in EF, draw FA, FG, F'D, FC, FH, FB : And because 
the two straight lines AE^ ED are equal to the two^BE, EC, 
and that they contain equal angles^ AED, BEC, the base AD, 
is equal'' to the base BC, and the angle DAE to the angle 
EBC : And the angle AEG is equal to the angle BEH"; there- 
fore the triangles AEG, BEH have two angles of one equal to 
two angles of the other, each to eaeh. and the sides AE, EB, 
adjacent to the equal angles, equal to onc another : where- 
fore they shall have their other sides equal"^ : GE is therefore 

equal 



OF EUCLID. 



^93 




b-i. 1. 



s. 1. 



vaual to EH, and AG to BH: and because AE is equal to ^°°^J^' 

EB, and FE common and at rightangles'to them, the base AF ^■'^'''^■^. 

is equal*" to the base FB ; for the same reast a, CF is equal to 

FD: And because AD is equal to BC, and AF to FB, the 

two sides FA, AD are equal to the two 

FB, BC, each to each j and the base p 

DF was proved equal to the base FC ; 

therefore the angle FAD is equal " to 

the angle FBC : Again, it was proved 

that GA is equal toBH, and also AF A.^ 

to FB ; FA, then, and AG, are equal 

to FB and BH, and the angle FAG 

has been proved equal to the angle 

FBH ; therefore the base GF is equal 

''to the base FH : Again, because it 

was proved, that GE is equal to EH, 

and EF is common ; GE, EF are 

equal to HE, EF ; and the base GF 

is equal to the base FH ; therefore the angle GEF is equal' 

to the angle KEF ; and consequently each of these angles is a 

right^ angle. Therefore FE makes right angles with GH, * lOdef. l. 

that is, with any straight line drawn through E in the plane 

passing through AB, CD. In like manner, it may be proved, 

that FE makes right angles with every straight line which 

meets it in that plane. But a straight line is at right angles 

to a plane when it makes right angles with every straight Tine , 

which meets it in that plane*": Therefore EF is at right an- 



gles to the plane in v/hich are AB, CD. 
straight line, 5cc. Q^ E D. 



Wherefore, if a 



' 3 d«i; 11. 



PROP. V. THEOR. 

JlF tliree straight lines meet all in one point, and aseex. 
straight line stands at right ang;les'to each of them 
in that point ; tliese three straight lines are in one 
and the same plane.' 

Let the straight line' AB stand at right angles to each of the 
straight lines BC, BD, BE, in B the point where they meet; 
BC, BD, BE, are in one and the same plane. 

If not, let, if it' be possible, BD and BE be in one plane, 
and BC be above it ; and let a plane pass through AB, BC, 
he common section of which v/ith the plane, in which BD 

O and 



19+ 



THE ELEMENTS 



M. II. 



Book x^, a^j BE are, shall be a straight*^ line ; let this be BF: There- 
^"sAL ^°^^ ^^^ three straight lines AB, BC, BF, are all in one plane, 

viz. that which passes through AB, BC; and because AB 

stands at right angles to each of the straight lines BD, BE, it 

is also at right angles ^ to the plane passing through them i 
^3 def. 11. ^pj therefore makes right angles'^ 

with every straight line meeting it -A. 

ill that plane ; but BF which is in 

that plane meets it : Therefore the 

angle ABF is a right angle ; but 

the angle ABC, by the hypothesis, 

is also a right angle ; therefore the 

angle ABF is equal to the angle 

ABC, and they are both in the 

same plane, which is impossible ; 

Therefore the straight line BC is 

not above the plane in which are BD and BE : Wherefore 

the three straight lines BC, BD, BE are in one and the same 

plane. Therefore, if three straight lines, &c. Q. E. D. 




PROP. VI. THEOR. 

1 F two straight lines be at right angles to the same 
plane, they shall be parallel to one another. 

Let the straight lines AB, CD be at right angles to the, 
same plane ; AB is parallel to CD. 

Let them meet the plane in the points B, D, and draw thij 
straight line BD, to which draw DE at right angles, in thel 
same plane; and make DE equal to AB, 
and join BE, AE, AD. Then, because 
AB is perpendicular to the plane, it 
*. 3 def. 11. shall make right' arjgles with every 
straight line which meets it, and is in 
that plane : But BD, BE, which are in 
that plane, do each of them meet AB. 
Therefore each of the angles ABD, 
ABE is a right angle : For the same rea- 
son, each of the angles CDB, CDE is 
a right angle : And because AB is equal 
to DE, and BD common, the two 
sides AB, BD are equal to the two 
ED, DB ; and tljey contain right angles ; therefore the base 
AD i» equal'' to thq base BE: Again, because AB is equal 




M. }. 



t« 



OF EUCLID. ' «95 

to DE,and BE to AD ; AB, BE are equal to ED, DA; and, ^;^*^- 

in the triangles ABE, EDA, the base AE is common : there- ^^"'■^^'^^ 

fore the angle ABE is equal <= to the angle EDA: But'8-i- 

ABE is a right angle : therefore EDA is also a right angle, 

and ED perpendicular to DA : But it is also perpendicular 

to each of the two ED, DC : Wherefore ED is at right angles 

to each of the three straight lines BD, DA, DC in the point in 

which they meet : Therefore these three straight lines are all 

in the same plane"^ : But AB is in the plane in which are BD, * ^- i^- 

DA, because any three straight lines which meet one another 

are in one plane*^ : Therefore AB, BD, DC are in one plane ; * 2. ij| 

And each of the angle's AED, BDC is a right angle; therefore 

AB is parallel^ to CD. Wherefore, if two straight lines, &c. '^^S. 1. 

Q,E. D. 

PROP. VII. THEOR. 

AF two straight lines be parallel, the straight line See2*» ' 
drawn from any point in the one to any point in 
the other, is in the same plane with the parallels. 

Let AB, CD be parallel straight lines, and take any point 
E in, the one, and the point F in the other: The straight line 
which joins E and F is in the same plane with the parallels. 

If not, let it be, if possible, above the plane, as EGF ; and 
in the plane ABCD in which the 

parallels are, draA^ the straight A "F "R 

line EHF from E to F ; and since - . . *M 

EGF also is a straight ime, the 
two straight lines EHF, EGF 
include a space between them, 

■which is impossibles There- \] »iOA.t. i. 

fore the straight line joining the C t D 

points E, F is not above the 

plane in which the parallels AB, CD are, and is therefore in 
that plane. Wherefore, if two straight lines, &c. Q. E. D. 



PROP.VIIL THEOR. 

XF two straight lines be parallel, and one of them set 
is at right angles to a plane ; the other also shall be 
at right angles to the same plane 

02 I^ 





JT'T. II. 



THEELEMENTS 

Let AB, CD be two parallel straight lines, and let one of 
them AB be at right angles to a ^lane ; the other CD is at 
right angles to the same plane. 

Let AB, CD meet the plane in the- points B, D, and join 
BD ; Therefore » ABj CD, BD are in one plane. In the plane 
to which AB is at right angles, draw DE at right angles 
to BD, and make DE equal to AB, and join BE, AE, AD. 
And because AB is perpendicular to the plane, it is perpen- 
dicular to every straight line which meets it, and is in that 

»3. del. 11. plane^ : Therefore each of the angles ABD, ABE, is a right 
angle : And because the straight line BD meets the parallel 
straight lines AB, CD, the angles ABD, CDB are''together 

'.'r>. 1. . aqual*" to two right angles: And ABD is a right angle; 
therefore also CDB is a right angle, and CD perpendicular to 
BD : And because AB is equal to DE, and BD common, the 
two AB, BD are equal to the two ED, 
DB, and the angle ABD is equal to j^^ 
the angle EDB, because each of them 
is a right angle ; therefore the" base AD 

' ' '• is equal <= to the base BE: Again, 
because AB is equal to DE, and BE to 
AD ; the two AB, BE, are equal to the 
two ED, DA j and the base AE is com- 
mon to the triangles ABE, EDA; 
wherefore the angle ABE is equal ^ to 
the angle EDA : And ABE is a right 
angle; and therefore EDA is a right 
angle, and ED perpendicular to DA: 
But it is also perpendicular to BD ; therefore ED is perpen- 
dicular'^ to the plane which passes through BD, DA, and shall*" 
make right angles with every straight line meeting it in that 
plane : But DC is in the plane passing through BD, DA, be- 
cause all three are in the plane in which are the parallels AB, 
CD : " Wherefore ED is at right angles to DC ; and therefore 
CD is at riglit angles to DE : But CD is also at right angles 
to DB ; CD then is at^ right "angles to. the two straight lines 
DE, DB in the point of theii* intersection D : and therefore 
is at right angles'" to the plane passing through DE, DB, which 
is the same plane to which A B is at right angles. Therefore, 
if two straight jjftes, &c. ^C^E..D. 



"8. 1, 



-4. n. 

'Jdef. 11 




OF EUCLID. 




PROP. IX. THEOR. 



X WO straight lines which are each of them paral- 
lel to the same straight line, and not in the same 
plane with it, are parallel to one another. 

Let AB, CD, be each of them parallel to EF, and not in the 
same plane with it; AB shall be parallel to CD, 

In EF cake any point G, from which draw, in the plane 
passing through EF, AB, the straight line GH at right angles 
to EF ; and in the plane passing through EF, CD, draw GK 
atrightanglesto the same EF. And 
because E^F is perpendicular both, 
to Gri and GK, EF is perpendi- 
cular* to the plane HGK passing 
through them : and EF is parallel 
toAB; therefore AB is at right 
angles *= to the plane HGK. For 
the same reason, CD is likewise at r? 
right angles to the plane HGK. 
Therefore AB, CD are each of them at right angles to the 
plane HGK. But if two straight lines are at right angles to 
the same plane, they shall be parallel*^ to one another. There- 
fore AB is parallel to CD. Wherefore, two straight lines, 
&c. Q, E. D. - 



A H 


rh 


\ 


J5 





F 



D 



»4. 11. 



8.11. 



6. n. 



PROP. X. THEOR. 



J F two straight lines meeting one another he pa- 
rallel to two others that meet one another, and are 
not in the same plane with the first two ; the first two 
and the other two shall -contain equal angles. 

•Let the two straight lines AB, BC, which meet one ano- 
ther, be paralleLto the two straight lines DE, EF that meet 
•ne another, and are not in the same plane with AB, BC. 
The angle ABC is equal to the angle DEF. 

Take BA, BC, ED, EF all equal to one another j and join 
O 3 AD, 



BookXI. 



'35. 1. 



og. 11. 

= l.Ax. 1. 



■is. 1. 



THE ELEMENTS 

AD, CF, BE, AC, DF : Because BA is equal and parallel to 
ED, therefore AD is* both equal and 
parallel to BE. For the same reason, 
CF is equal and parallel to BE. There- 
fore AD and CF are each of them 
equal and parallel to BE. But straight 
lines that are parallel to the same straight 
line, and not in the same plane with it, 
are parallel'' to one another. Therefore 
AD is parallel to CF ; and it is equal*: 
to it, and AC, DF join them towards 
the same parts ; and therefore ' AC is 
equal and parallel to DF. And be- 
cause AB, BC are equal to DE, EF, and the base AC to the 
base DF ; the angle ABC is equaH to the angle DEF.. 
Therefore, if two straight lines, &c. Q. E. D. 




*12. 1. 



bli. I. 



'31. 1. 



'4. 11. 



«8. 11. 



*3def. 11. 



PROP^^I. PROB. 

1 O draw a straight line perpendicular to a plane, 
from a given point above it. 

Let A be the given point above the plane BH ; it is required 
to draw from the point A a straight line perpendicular to the 
plane BH. 

In the plane draw any straight line BC, and from the point 
A draw ^ AD perpendicular to BC. If then AD be also per- 
pendicular to the plane BH, the thing required is already 
done ; but if it be not, from the 
point Ddraw'*, in the plane BH, 
the straipjht line DE, at right an- 
gles to BC ; and from the point 
A draw AF perpendicular to 
DE ; and through F draw"^ GH 
parallel to BC : And because BC 
is at right angles to ED and DA, 
BC is at right angles'* to the plane 
passing through ED, DA. And 
GH is parallel to RC ; but, if 
two straight lines be parallel, one of which is at right angles to 
a plane, the other shall be at right <^ angles to the same plane ; 
wherefore GH is at right angles tothe plane through ED,DA, 
and is perpendicular ^ to every straight line meeting it in that 
plane. But AF, which is in the plane through ED, AD, meets 

it: 




OF EUCLID. ,99 

it : Therefore GH Is perpendicular to AF ; and consequently ^ook XI. 
AF is perpendicular to GH ; and AF is perpendicular to DE: ^*^V"**^ 
Therefore AF is perpendicular to each of the straight lines GH, 
DE. But if a straight line stands at right angles to each of 
two straight lines in the point of their interseclionj i: shall also 
be at right angles to the plane passing through them. But the 
plane passing through ED, GH is the plane BH ; therefore 
AF is perpendicular to the plane BH ; therefore, from the given 
point A, above the plane BH, the straight line AF is drawn 
perpendicular to that plane : Which was to be done. 



PROP. XII. PROB. 

X O erect a straight line at right angles to a given 
plane, from a point given in the plane. 

Let A be the point given in the plane ; it is required re£t 
a straight line from the point A at right ^^ ^ 



angles to the plane. -^ 

From any point B above the plane 
draw* BC perpendicular to it ; and 
from A draw^ AD parallel to BC. 
Because, therefore, AD, CB are two 
parallel straight lines, and one of them 



B 



*ii. 11. 

»31. U 



BC is at right angles to the given plane, / j\ f! / 

the other AD is also at right angles to 

it^ Therefore a straight line has been erected at right angks c g. ^ 
to a given plane from a point given in it. Which was to be 
done. 



PROP. Xm. THEOR. 

-T ROM the same point in a given plane, there cannot 
be two straight lines at right angles to the plane, 
Bpon the same side of it ; and there can be but one per- 
pendicular to a plane from a point above the plane. 

For, if it be possible, let the two straight lines AB, AC, be 
at right angles to a given plane from the same point A in» the 
plane, and upon the same side of it; and let a plane pass through 
B Aj AC i the common section of this with the given plane is a 

O 4 straight 



200 
Book XI. 



^S. U. 



6. 11. 



THE ELEMENTS 

straight* line passing through A: Let DAE be their common 
se£lion: Therefore the straight lines AB, AC, DAE are in one 
plane : And because CA is at right angles to the given plane, it 
shall make right angles with every 
straight line meetingit inthatplane. 
But DAE, which is in that plane, 
meets CA ; therefore CAE is a right 
angle. For the same reason BAE 
is a right angle. Wherefore the an- 
gle CAE is equal to the angle 
BAE; and they are in one plane, 
■which is impossible. Also, from a point above a plane, there 
can be but one perpendicular to that plane ; for, if there could 
be two, they would be parallel'' to one another, which is ab- 
surd. Therefore, from the same point, &c. Q. E. D. 




PROP. XIV. THEOR. 



■■3M. 1 



«• 17. 1. 



'S(ief. U. 



1 LANES to which tli^ same straight line is per- 
pendicular, are parallel to one another. 

Let the straight line AB be perpendicular to each of the 
planes CD, EP' s these planes are parallel to one another. 

If not, they shall meet one another when produced ; let them 
meet ; their common sed"ion shall be 
a straight line GH, in which take any 
point K, and join AK, BK : Then, 
because AB is perpendicular to the 
plane EF, it is perpendicular" to the 
straight line BK which is in that plane. 
Therefore ABK is a right angle. For 
the same reason BAK is a right angle ; 
wherefore the two angfes ABK, BAK 
of the triangle ABK are equal to two 
right, angles, which is impossible'' ; 
Therefore the planes CD, EP\ though 
produced, do not meet one another j 
that is, they are parallel. There- 
fore planes, &c. Q. K. D. 




OF EUCLID. 



PROP. XV. THEOR. 




51. I. 



«3dctll. 



jlF two straiglil lines meeting one another, be pa- sceN. 
rallel to two straight lines which meet one another, 
but are not in the same plane with the first two ; the 
plane M-hich passes through these is parallel to t he 
plane passing through the others. 

Let AB, BC, two straight lines meeting one another, be 
parallel to DE, EF that meet one another, but are not in the 
same plane with i\B, BC : The planes through AB, BC, and 
D£, EF shall not meet, though produced. 

From the point B draw BG perpendicular* to the plane * n. n. 
which passes through DE, EF, and let it meet that plane in 
Gj and through G draw GH parallel'' to ED, and GK 
parallel to EF : And because BG is perpendicular to the plane 
through DE, EF, it shall 
make right angles with every 
straight line meeting it in that 
planed But the straight lines 
GH, GK ill that plane meet 
it: Therefore each of the 
angles BGH, BGK is a right 
angle : And because BA is 
parallel'' to GH (for each of 
them is parallel to DE, and 
they are rtot both in the same plane with it) the angles GBA, 
BGH are together equa^ to two right angles : And BGH is ' 29, i. 
a right angle ; therefore also GBA is a right angle, and GB 
perpendicular to BA : For the same reason, GB is perpendi- 
cular to BC : Since therefore the straight line GB stands at 
right angles to the two straight lines BA, BC, that cut one 
another in B ; GB is perpendicular* to the plane through BA, f^ ij_ 
BC : And it is perpendicular to the plane through DE, EF j 
therefore BG is perpendicular to each of theplanes through AB, 

8C, and DE, EF : But planes to which the same straight line 
perpendicular, are parallel? to one another : Therefore the g f^, n 
plane through AB, BC is parallel to the plane through DE, 
EF. Wherefore if two straight lines, &c. Q. E. D. 





See N. 



THE ELEMENTS 



PROP. XVI. THEOR. 

JlT two parallel planes be cut by another plane, 
their common sections with it are parallels. 

Let the parallel planes, AB, CD be cut by the plane EFHG, 
and let their common sections with it be EF, GH : EF is 
parallel to GH. 

For, if it is not, EF, GH shall meet, if produced, either on 
thesideof FHjOr EG : First, let them be produced on the side 
of FH, and meet in the point K : Therefore, since EFK is in 
the plane AB, every point in 
EFK is in that plane ; and K 
^3 a point in EFK ; therefore 
K is in the plane AB : For 
the same reason K is also in 
the plane CD : Wherefore the 
planes, AB, CD, produced 
meet one another ; but they 
do not meet, since they are 
parallel by the hypothesis : 
Therefore, the straight lines 
EF, GH do not meet when 
produced on the side of FH : In the same manner it may be 
proved, that EF, GH do not meet when produced on the side 
of EG : But straight lines which are in the same plane, and do 
not meet, though produced either way, are parallel : There- 
fore EF is parallel to GH. Wherefore, if two parallel planes, 
&c. Q. E. D. 




D 



PROP. XVn. TIIEOR. 

If two straight lines be cut by parallel planes, 
they shall be cut in the same ratio. 

Let the straight lines AB, CD be cut by the parallel planes 
GH, KL, MN, in the points A, E, B j C, F, D : As AE Hi 
to EB, so is CF to FD. 

Join AC, BD, AD, and let AD meet the plane KL in th< 
point X: and join EX, XF: Because the two parallel plane^ 
KL, MN are cut by the plane EBDX, the common se^ion? 

EX, 



OF EUCLID. 



203 



EX, BD are prirallels For the same reason, because the two ^*«^ ^ 

parallel planes GH, KL are cut 1 }tj_ j 1^ 

by the plane AXFC, the com- ^__ \\ 

mon sections AC, X Fare parai- / a 

lel : And because EX is paral- qL — L- 

lel to BD, a side of the triangle 

ABD, as AE to EB, so is^ AX 

to XD. Again, because XF is 

parallel to AC, a side of the 

triangle ADC, as AX to XD, 

so is CF to FD: And it was ^^ 

proved that AX is to XD, as 

AE to EB : Therefore% as I ^^_\]^ xr MI.5. 

AE to EB, so is CF to FD. 

V/herefore, iftwostraightlines, \\^ 

&c. Q. E. D. 




PROP. XVIII. THEOR. 



F a straight line be at right aagles to a plane, every 
plane which passes throu^ it shall Jjje at right an- 
gles to that jjlane. 

Let the straight line AB be at right angles to a plane CK ; 
every plane which passes chrovgh AB shall be at right angles 
to the plane CK. 

Let any plane DE pass through AB, and let CE be the com- 
I mon section of the planes DE, CK; take any point F in CE, 
'rn which draw FG in the 

-ae DE at right angles to 
CE : And because AB is per- 
pendicular to the plane CK, 
therefore it is also perpendi- 
cular to every straight line in 

that plane meeting it^: And \ \ \ »3def, 11, 

consequently it is perpendicu- 
lar to CE : Wherefore ABF 
is a right angle ; But GFB is 



■ 


D G A Tf 








K 


\ 






\ 


\ 






\ 



F B 



E 

likewise a right angle ; therefore AB is parallel'' to FG. And b og. j, 
AB is at right angles to the plane CK ; therefore FG is also at 
right angles to tiie same planed But one plane is at right an- c 3. n. 
js to another plane when the straight lines drawn in one of the 
-nes, at right angles to their common section, are also at right 

angles 



,204 



THE ELEMENTS 



Book XI. anglcs to the other plane' ; and any straight line FG in the 
d i def. 11. plane DE, which is at right angles to CE the common sedtion 
of the planes, has been proved to be perpendicular to the other 
plane CK ; therefore the plane DE is at right angles to the 
plane CK. In like manner, it may be proved that all the planes 
which pass through AB are at right angles to the plane CK. 
Therefore, if a straight line. Sec. Q^ E. D. 



PROP. XIX. THEOR. 



XF two planes cutting one another be each of them 
perpendicular to a thud plane ; their common sec- 
tion shall be perpendicular to tlie same plane. 

Let the two planes AB, BC be each of them perpendicular 
to a third plane, and let BD be the commen section of the first 
two ; BD is perpendicular to the third plane. 

If it be not, from the point D draw, in the plane AB, the 
straight line DE at right angles to AD the common se6lion of 
the plane AB with the third plane ; and in the plane BC draw 
DF at right angles to CD the common sedlion of the plane BC 
with the third'plane. And because the 
plane AB is perpendicular to the third 
plane, and DE is drawn in the plar.e 
AB at right angles to AD their com- 
mon section, DE is perpendicular to 
+ def. 11. the third plane*. In the same manner, 
• it may be proved that DF is perpen- 
the third plane. Where- 
the point D two straight 
at right angles to the third 
same side of it, which 
Therefore, from the 



fci3. n. 



dicular to 
fore, from 
lines stand 
plane, upon the 
is impossible'' 




point D there cannot be any straight 

line at right angles to the third plane, 

except BD the common section of the planes AB, BC. Bl 

therefore is perpendicular to the third plane. Wherefore^ ii 

two planes, &c. Q. E. D. 



OF EUCLID. 205 

Bock XI. 

PROP. XX. THEOR. "^'^^ 

I Fa solid angle be contained by three plane an-'s« n. 
gles, any two of them are greater than the third. 

Let the solid angle at A be contained by the three plane 
angles, BAG, CAD, DAB. Any two of them are greater 
than the third. 

it' the angles BAC, CAD, DAB be all equal, it is evident 
that any two of them are greater than the third. But if they 
are not, let BAC be that angle which is not less than either of 
the other two, and is greater than one of them DAB ; and at 
the point A iij the straight line iVB, make, in the plane which 
passes through BA, AC, the angle BAE equal^ to the angle »23. 1. 
DAB ; and make AE equal to AD, and through E draw 
EEC cutting AB, AC in the points 
B, C, and join DB, DC. And be- 
cause DA is equal to AE, and AB is 
common, the two DA, AB are equal 
to the two EA, AB, and the angle 




'Dab is equal to the angle EAB : 

Therefore the base DB is equal* to / --^" "^^^^^^^^^-^^ — \ b-i. 1. 
the base BE. And because BD, DC 
are greater "= than CB, and one of "*' "^^ ^ c 00. 1. 

them BD has been proved equal to BE a part of CB, therefore 
the other DC is greater than the remaining part EC. And be- 
cause DA is equal to AE, and AC common, but the base DC 
greater, than the base EC ; therefore the angle DAC is greater** <i 25. 1. 
than the angle E AC ; and, by the constru£tion, the angle DAB 
jis equal to the angle BAE ;* wherefore the angles DAB, DAC 
I are together greater than BAE, EAC, that is, than the angle 
I BAC. But^BAC is noMess than either of the angles DAB, 
iDAC ; therefore BAC, with either of them, is greater than 
'the other. Wherefore if a solid angle, &c. Q. E. D. 

PROP. XXL THEOR. 

^-tliVERY solid anajie is contained bv plane auo-les, 
[which together are less than four right angles. 

First, Let the solid angle at A be contained by three plane 
jangles BAC, CAD, DAB. These three together are less 
ithan four right angles. 

? Take 



206 

Book XI. 



^'QO. 11. 



> 32. 1. 




THE ELEMENTS 

Take in each of the straight lines AB, AC, AD any points 
B, C, D, and join BC, CD, DB : Then, because the solid angle 
at B is contained by the three plane angles CBA, ABD, DBG, 
any two of them are greater^ than the third ; therefore the 
angles CBA, ABD are greater than the angle DBC: For the 
same reason, the angles BCA, ACD are greater than the angle 
DCB ; and the angles CDA, ADB greater than BDC : Where- 
fore the six angles CBA, ABD, JiCA, ACD, CDA, ADB 
are greater than the three angles DBC, 
BCD, CDB : But the three angles 
DBC, BCD, CDB are equal to two 
right angles'' : Therefore the six angles 
CBA, ABD, BCA, ACD, CDA, 
ADB are greater than two right angles : 
And because the three angles of each 
of the triangles ABC, ACD, ADB are j>~ 
equal to two right angles, therefore the 
nine angles of these three triangles, viz. the angles CBA, 
BAC,ACB,ACD, CDA, DAC, ADB, DBA, BAD are 
equal to six right anf^les: Of these the six angles CBA, 
ACB, ACD, CDA, ADB, DBA are greater than two right 
angles : Therefore the remaining three angles BAC, DAC, 
BAD, which contain the solid angle at A, are less than four 
right angles. • 

Next, Let the solid angle at A be contained by any number 
of plane angles BAC, CAD, DAK, EAF, FAB : these 
together are less than four right angles. 

Let the planes in which the angles are, be cut by a plane, and 
let the common sections of it with those 
planes be BC, CD, DE, EF, FB : And 
because the solid angle at B is contain- 
ed by three plane angles CBA, ABF, 
FBC, of which any two are greater^^ 
than the third, the angles CBA, ABF 
are greater than the angle FBC : for 
the same reason, the two plane angles 
at each of the points C, D, E, F, viz. 
the angles which are at the bases of the 
triangles having the common vertex 
A, are greater than the third angle at 
the same point, which is one of the an- 
gles of the polygon BCDEF : There- 
fore all the angles at the bases of the triangles are together. 

greater 




OF EUCLID. 207 

greater than all the angles of the polygon : And becausa all the Book xi. 
angles of the triangles are together equal to twice as many ""^^ 
right angles as there are triangles^ ; that is, as there are sides » Z2. 1. 
in the polygon BCDEF ; and that all the angles of the poly- 
gon, together with four right angles, are likewise equal to 
twice as many right angles as there are sides in the polygon*^ ; * '• ^°''* 
therefore all the angles of the triangles are equal to all the 
angles of the polygon together with four right angles. But all 
the angles at the bases of the triangles are greater than all the 
angles of the polygon, as has been proved. Wherefore the 
remaining angles of the triangles, viz. those at the vertex, 
which contain the solid angle at A, are less than four right an- 
gles. Therefore every soUd angle, &c. Q. E. D. 



PROP. XXII. THEOR. 

JlF every two of three plane angles be greater than s^^"- 
the third, and if the straight lines which contain 
tliem be all equal ; a triangle may be made of the 
straight lines that join the extremities of those equal 
straight lines. 

Let ABC, DEF, GHK be three plane angles, whereof 
every two are greater than the third, and are contained bv the 
equal straight lines AB, BC, DE, EF, GH, HK ; if their 
extremities be joined by the straight lines AC, DF, GK, a tri- 
angle may be made of three straight lines equal to AC, DF, 
GK ; that is, every two of them are together greater than the 
third. 

If the angles at B, E, H, are equal, AC, DF, GK are also 
(equal* and any two of them greater than the third : But if**- ^• 
the angles are not all equal, let the angle ABC be not less than 
jeither of the two at E, H ; therefore the straight line AC is 
not less than either of the other two DF, GK" 3 and it is " 4. Cor. 
plain that AC, together with either of the other two, must be ^*' '• 
greater than the third : Also DF with GK are greater than 
AC : For, at the point B, in the straight line AB, make'' the <^q3. 1. 

ande 



o8 THE ELEMENTS 

Book XI. angle ABL equal to the angle GHK, and make BL equal to 
^-^'^^ one of the straight lines AB, BC, DE, EF, GH, HK, and join 
AL, LC ; Then, because AB, BL are equal to GH, HK, and 
the angle ABL to the angle GHK, the base AL is equal to 
the base GK : And because the angles at E, Hare greater than, 
the angle ABC, of which the angle at H is equal to ABL, there- 
fore the remaining angle at E, is greater thau the angle LBC : 



ScfN. 





F G 



And because the two sides LB, BC are equal to, the two DE, 
EF, and that the angle DEE is greater than the angle LBC, the 

* 24. 1. base DF is greater'' than the base LC: And it has been proved 
that GK is equal to AL ; therefore DF and GK are greater 

' 20. 1. than AL and LG: But AL and LC are greater^ than AC j much 
more then areDF andGKgreater than AC. Whereroreevciy 
two of these straight lines AC, DF, GK are greater than the 

•^32.1. third; and, therefore, a triangle may be made*^, the sides of 
which shall he equal to AC, DF, GK. Q. E. D. 



PROP. XXIIL PROB. 



X O make a solid angle which shall be contained 
by three given plane angles, any two of them being 
ii-reater than the third, and all three too'Cther less 
than four riglit angles. 






Let the three given plane angles be ABC, DEF, GHK,' 
any two of which are greater than the third, and all of thenif 
together less then four right angles. It is required to make a 'V 
solid angle contained by three plane angles equal to ABC, * 
DEF, GHK, each to each. 

. - From 



OF EUCLID. 



209 



From the straight lines containing the angles, cut ofF AB ; Book XI. 
BC, DE, EF, Gil. HK, all equal to one another j and join ""^'"^ 
AC, DFj GK : Then a triangle may be made' of three straight » 22. 11. 




22. 1. 
5. 4. 



lines equal to AC, DF, GK. Let this be the triangle 
LMN", so that AC be equal to LM, DF to MN, and GK - 
to LN i and about the triangle LMN describe*^ a circle, and c 
find its centre X, which will either be within the triangle, or 
in one of its sides, or without it. 

First, Let the centre X be within the triangle, and join 

LX, MX, NX : AB is greater than LX : If not, AB must 

either be equal to, or less than LX ; first, let be equal : Then 

because AB is equal to LX, and that AB is also equal to BC, 

and LX to XM, AB and BC are equal to LX and XM, each 

to each ; and the base AC is by construdlion, equal to the 

base LM 3 wherefore the angle ABC is equal to the angle 

I LXM**. For the same reason, the angle DEF is equal to the <i 8. 1. 

langle MXN, and the angle GHK 

to the angle NXI, : Therefore the 

three angles ABC, DEF, GHK are 

(equal to the three angles LXM, 

jMXN, NXL: But the three angles 

[LXM, MXN, NXL are equal to 

Ifour right angles'^; therefore also 

Oie three andes ABC, DEF, GHK 

ir^equal to four right angles : But, 

>y the hypothesis, they are less vf^ 
lan four right angles, which is 

absurd j therefore AB is not equal 

to LX: But neither can AB be 

ess than LX : For, if possible, let it be less, and upon the 
raight line LM, on the side of it on which is the centre X, 
-scribe the triangle LOM, the sides LO, OM of which are 
iual to AB, BC ; aud because the base LM is equal to the 

P base 




2IO 



THE ELEMENTS 




^°^^^- based AC, the angle LOM is equal to the angle ABC^ : And 

4 sPfT*'^ ■A-Bj *^^^ ^^' LO, by the hypothesis, is less than LX; where- 
fore LO, OM tail within the triangle LXAl j for, if they fell 
upon its sides, or without it, they 
would be equal to or greater than 

^21.1. LX, XM^: Therefore the angle 
LOM, that is, the angle ABC, is 
greater than the angle LXM^: In 
the same manner it may be proved 
that the angle DEF is greater than 
the angle MXN, and the angle 
GHK greater than the angle NXL. 
Therefore the three angles ABC, 
DEF, GHK are greater than the 
three angles LXM, MXN, NXL ; 
that is, than four right angles : But 
the same angles ABC, DEF, GHK are less than four right 
angles i which is absurd : Therefore AB is nofless than LX, 
and it has been proved that it is not equal to LX ; wherefore 
AB is greater than LX. 

Next, Let the centre X of the circle fall in one of the sides 
of the triangle, viz. in MN, and » 
join XL : In this case also AB is 
greater than LX. If not, AB is 
either equal to XL, or less than it : 
First, ]?t it be equal to XL : There- 
fore AB and BC, that is, DE, and 
EF, are equal to MX and XL, that 
iSjtoMN: Butjby the construction, 
MN is equal toDF; therefore DE, M'f 
EF are equal to DF, which is im- 

t20. 1. possiblef: Wherefore A B is not 
, equal to LX; nor is it less; for then, 
much more, an absurdity would 
follow : Therefore AB is greater than LX. 

But, let the centre X of the circle fail without the triangle 
LMN, and join LX, MX, NX. In this case likewise AB is i 
greater than LX : If not, it is either equal to, or less than LX: j 
First, let it be equal ; it may be proved in the same manner, i, 
as in the first case, that the angle ABC is equal to the angle 
MXL,and GHK to LXN; therefore the whole angle MX>f is 
equal to the two angles, ABC, GHK : But ABC and GHK 
are together greater than the angle DEF ; therefore also 
the angle MXN is greater than DEF, And because DE, 

EF 




OF EUCLID. 



211 



EF are equal to MX, XN, and the base DF to the base ^ook. xi- 
MN, the angle MXN is equal"^ to the ang]e DEF : And it has ^^f*^ 
been proved, that it is greater than DEF, which is absurd. 
Therefore A B is not equal to LX. Nor yet is it hss ; for then, 
as has been proved in the first case, the angle ABC is greater 
than the angle MXL, and the angle GHK greater than the 
angle LXN . At the point B, in the straight lineCB,make the 
angle CBP equal to the angle GHK, and make BP equal to 




/ 






/ 

G~ 






HK, and join CP, AP. And because CB is equal to GH ; 
CB, BP are equal to GH, HK, each to each, and they contain 
equal angles; wherefore the base CP is equal to the base GK, 
that is, to LN. And in the isosceles triangles ABC, MXL, 
because the angle ABC is greater than the angle MXL, there- 
fore the angle MLX at the base is greater 5 than the angle t sg 
AC Bat the base. For the same reason,because the angle GHK, 
or CBP, is greater than the angle 
LXN, the angle XLN is greater 
than the angle BP. Therefore the 
whole angle MLX is greater than 
the whole angle ACP. And because 
ML, LN are equal to AC, CP, 
each to each, but the angle MLN 
is greater than the angle ACP, the 
base MN is greater ^ than the base 
AP. And MN is equal to DF ; 
therefore also DF is greater than 
AP. Again, because DE, EF are 
equal to AB, BP, but the baseDF 
greater than the base AP,the angle 
DEF is greater'' than the angle 

ABP. And ABP is equal to the two angles ABC, CB", that 
IS, to the two angles ABC, GHK ; therefore the angle IJEF is 
greater than the two angles ABC, GHK ; "but it is also less 
than these, which is impossible. Therefore AB h not less than 

P2 LX; 




»24. 1. 



2J. I. 



212 



THE ELEMENTS 



Book XI. LX, and it has been proved that it is not equal to it ; there- 

^"^^^''^^ fore AB is greater than LX. 

« 12. 11. From the point X ere6l* XR at right angles to the plane 
of the circle LMN. And because 't has been proved in all the 
cases, that AB is greater than LX, find a square equal to the 
excess of the square of AB above 
the square of LX, and make HX 
"* equal to its side, and join RL,RM, 

RN. Because RX is perpendicular 
to the plane of the circle LMN, it 

" Sdef. 11, is'' perpendicular to each of the 
straight lines LX, MX, NX And 
because LX is equal to MX, and 
XR common, and at right angles 
to each of tbem, the base RL is 
equal to the base RM. For the same 
reason, RN is equal to each of the 
twoRLjRM. 1 hereforethethree 
straight lines RL,RM,RN,areaIl 
equal. And becausp the square of 

XR is equal to the excess of the square of AB above the square 
of LX ; therefore the square of AB is equal to the squares of 

>- 47, 1. LX, XR. But the square of RL is equal' to the same squares, 
because LXR is a right angle. Therefore the square of AB 
is equal to the square of RL, and the straight line AB to RL. 
But each of the straight lines BC, DE, EF, GH, HK is equal 
to AB, and each of the twoRM,RN is equal to RL. Where- 
fore AB, BC, DE, EF, GH, HK, are each of them equal to 
each of the straight lines RL, RM, RN. And because RTv, 
RM, are equal to AB, BC, and the base LM to the base AC ; 

« 8. 1, the angle LRM is equal*^ to the angle ABC. For the same 
reason, the angle MRN is equal to the angle DEF, and NRL 
to GHK. Therefore there is made a solid angle at R, which 
is contained by three plane angles LRM, MRN, NRL, which 
are equal to the three given plane angles ABC, DEF, GHK, 
each to each. Which was to de done. 





OF EUCLID. 



PROP. A. THEOR. 

If each of two solid angles be contained by three s^ex. 
plane angles equal to one another, each to each ; 
the planes in which the equal angles are to have the 
same incUnation to one another. 

Let there be two solid angles at the points A, B ; and let 
the ano-le at A be contained by the three plane angles CAD, 
CAE,^AD ; and the angle at B by the three plane angles 
FBG FBH HBG ; of which the angle CAD is equal to the 
angle FBG,' and CAE to FBH, and EAD to HBG : The 
planes in which the equal angles are, have the same inclination 
to one another. 

In the straight line AC take any point K, and in the plane 
CAD from K draw the straight line KD at right angles to AC, 
and in the plane CAE 
the straight line KL A. 

at right angles to the 
same AC : Therefore 
the angle DKL is the 
inclination * of the 
plane CAD to the 
plane CAE: In BF 
take BiVl equal to AK, 
and from the point M 

draw, in the planes FBG, FBH, the straight lines MG, MN 
at right angles to BF ; therefore the angle GMN is the incli- 
nation ^ of the plane FBG to the plane FBH : Join LD, NG; 
and because in the triangles KAD, MBG, the angles KAD, 
MBG are equal, as also the right angles AKD, BMG, and 
that the sides AK, BM, adjacent to the equal angles, are equal 
to one another j therefore KD is equal'' to AIG, and AD to 
BG: For the same reason, in the triangles KAL, MBN, 
KL is equal to MN, and AL to BN : And in the triangles 
LAD, NBG, LA, AD are equal to NB, BG, and they con- 
tain equal angles: therefore the base LD is equal' to the '4.1: 
base NG. Lastly, in the triangles KLD, MNG, the sides 
DK, KL, are equal to GM, MN, and the base LD to the 
base NG ; therefore the angle DKL is equal to"* the angle * 8. i. 
GMN : But the angle DKL is the inclination of the plane 
CAD to the plane CAE, and the angle GMN is the inclina- 

P 3 tion 




»6.de£.ll. 



26. 1. 



214 ' ' *^HZ ELEMENTS 

Book XI. tion of the, plane FBG to the plane FBH, which planes have 

»7*dcfn therefore the same inclination^ to one another: And in the 

' sanme manner it may be demonstrated, that the other planes in 

w;hich the equal angles are, have the same inclination to one 

another. Therefore, if two solid angles, &c. Q^ E. D. 



PROP. B. THEOR. 

seeN. If two soUd aiiglcs bc contained, each by three 
plane angles which are equal to one another, each to 
each, and alike situated ; these solid angles are equal 
to one another. 

Let there be two solid angles at A and B, of which the solid 
angle at A is contained by the three plane angles CAD, CAE, 
EAD : and that at B, by the three plane angles FBG, FBH, 
HBG; of which CAD is equal to FBG; CAE to FBH; 
and EAD to HBG : The solid angle at A is equal to the solid 
angle at B. 

Let the solid angle at A be af)plied to the solid angle at B ; 
and, first, the plane angle CAD being applied to the plane 
angle FBG, so as the point A may coincide with the point B, 
and the straight line AC with BFj then AD coincides with 
BG, because the angle CAD 
is equal to the angle FBG : 
And because the inclination of 

^ the plane CAE to the plane 

^'' CAD is equal => to the inclina- 
tion of the plane FBH to the 
plane FBG, the plane CAE 
coincides with the plane FBH, 
because the planes CAD, FBG coincide with one another : 
And because the straight lines AC, BF coincide, and that the 
angle CAE is equal to the angle FBH ; therefore AE coin- 
cides with BH, and AD coincides with BG ; wherefore the 
plane EAD coincides with the plane HBG : Therefore the 
solid angle A coincides with the solid an2;le B, and conse- 

» 8. A. 1. quently they are equal ^ to one another. Q. E. D. 





OF EUCLID. 



PROP. C. THEOR. 

OOLID figures contained by the same number ofseeN. 
equal and similar planes alike situated, and having 
none of their solid angles contained by more than 
three plane angles ; are equal and similar to ©ne ano- 
ther. 

Let AG, KQ^be two solid figures contained by the same 
number of similar and equal planes, alike situated, viz. let the 
plana AC be similar and equal to the plane KM, the plane 
AF to KP ; BG to LQ_; GD to QN -, DE to NO; and, 
lastly, FH similar and equal to PR : The solid figure AG is 
equal and similar to the solid figure KQ^ 

Because the solid angle at A is contained by the three plane 
angles BAD, B AE, EAD, which, by the hypothesis, are equal 
to the plane angles LKiN, LKO, OKN, which contain the 
solid angle at K, each to each j therefore the solid angle at A 
is equal'' to the solid angle at K: In the same manner, the :. . 
other solid angles of the figures are equal to one another. If, 
then, the solid figure AG be applied to the solid figure KQ_, 
first the plane 



4 G 


\E 


\ 


■ 


^ 


\! 



.0 



Q. 

1\ 



^ 



M 



K 



figure AC being 
applied to the 
plane figure KM ; 
the straight line 

AB coinciding DV [ 's;;' N 

with KL, the 

figure AC must -^- B 

coincide with the 

figure KM, because they are equal and similar : Therefore the 
straight lines AD, DC, CB coincide with KN, NM, ML, 
each with each ; and the points A, D, C, B, with the points 
K, N, M, L: And the solid angle at A coincides with* the 
solid angle at K ; wherefore the plane AF coincides with the 
plane KP, and the figure AF with the figure KP, because they 
are equal and similar to one another : Therefore the straight 
lines AE, EF, FB, coincide with KO, OP, PL; and the 
points EF with the points O, P. In the same manner, the 
figure AH coincides with the figure KR, and the straight line 
DH with NR, and the point H with the point R: And be- 
cause the solid angle at B is equal to the solid angle at L, it 
may be proved, in the same manner, that the figure BG coin- 

P 4 cide« 



2l6 



THE ELEMENTS 



Book XI. cides with the figure LQ, and the straight line CG with MQ, 
^^''^"^ and the point G with the point Q: Since, therefore, all the 
planes and sides of the solid figure AG coincide with the 
planes and sides of the solid figure KQ, AG is equal and si- 
milar to KQ^: And, in the same manner, any other solid fi- 
gures whatever contained by the same number of equal and 
similar planes, alike situated, and having none of their solid an- 
gles contained by more than three plane angles, may be proved 
to be equal and similar to one another. Q^ E. D. 



PROP. XXIV. THEOR. 



SeeN. 



• 16. 11. 



"10. 11. 



•=4. 1. 
"3*. 1. 



XF a solid be containecl by six planes, two and t\v@ 
of which are parallel ; tlie opposite planes are simi- 
lar and equal parallelograriis. 

Let the solid CDGH be contained by the parallel planes 
AC, GF ; BG, CE ; FB, AE : Its opposite planes are si- 
milar and equal parallelograms. 

Because the two parallel planes BG, CE, are cut by the 
plane AC, their common sections AB, CD are parallel*. 
Again, because the two parallel planes BF, AE are cut by the 
plane AC, their common sections AD, BC are parallel^ : And 
AB is parallel to CD ; therefore AC is a parallelogram. In 
like manner, it may be proved that 
each of the figures CE, FG, GB, 
BF. AE is a parallelogram : Join 
AH, DF; and because AB is paral- 
lel to DC, and BH to CF ; the two 
straight lines AB, BH, which meet 
one another, are parallel to DC 
and CF, which meet one another, 
and are not in the same plane with 
the other two ; wherefore they con- 
tain equal angles'' ; the angle ABH is therefore equal to the 
angle DCF : And because AB, BH arc equal to DC, CF, and 
the angle ABH equal to the angle DCF ; therefore the base 
AH is equal "^ to the base DF, and the triangle ABH to the 
triangle DCF : And the parallelogram BG is double^ of the 
triangle ABH, and the parallelogram CE double of the triangle 
DCF ; therefore the parallelogram BG is equal and si. ilar 
to the parallelogram CE. In the same manner it may be 
proved, that the parallelogram AC is equal and similar 

to 




OF EUCLID. 



217 



to the parallelogram GF, and the parallelogram AE to BF. BookXI. 
Therefore, if a solid, &c. Q, E. D. ^-^-^ 



PROP. XXV. THfeOR. 



XF a solid parallelopiped be cut by a plane parallel to ^**^' 
two of its opposite planes ; it divides the wiiole into 
two solids, the base of one of which shall be to the 
base of the other, as the one solid is to the other. 

Let the solid parallelopiped ABCD be cut by the plane 
EV, which is parallel to the opposite planes AR, HD,and di- 
vides the whole into the two solids ABFV, EGCD'j as the 
base AEFY of the first is to the base EHCF of the other, so 
is the solid ABFV to the solid EGCD. 

Produce AH both ways, and talce any number of straight 
lines HM, MN, each equal to EH, and any number AK, KL, 
each equal to EA, and complete the parallelograms LO, KY, 
HQ,, MS, and the solids Li', KR, HU, MT: Then, because 
thestraightlines LK, KA,AE are all equal, the parallelograms 



X 



L 



Z 



ES 



H 



C 



lis. 



\Y 



O 



Y 



J) 



H 



^^M-4 



SN^T 



M_ 



\LAi 



iN 



C "0. s 



LO, KY,AF are equal*: And likewise the parallelograms KX,»*^- ^• 
KB, AG* ; as also ^ the parallelograms LZ, KP, AR, because " -•*• ^' 
they are opposite planes : For the same reason, the parallelo- 
grams EC, HQ^, MS, are equaP ; and the parallelograms HG, 
HI, IN, as also^ HD, MU, NT : Therefore three planes of 
the solid L P are equal and similar to th ree planes of the solid K R, 
as also to three planes of the solid AV : But the three planes 
opposite to these three are equal and similar ^ to them in the 
several solids, and none of their solid angles are contained by 
more than three plane angles : Therefore the three solids LP, 
KR, AV are equal'= to one another: For the same reason the « c. i! 
three solids ED, HU, MT are equal to one a:iother : There- 
fore 



2l8 



THE ELEMENTS 



Book XI. (q^q what multiple soever the base LF is of the base AF, the 
'^"^'"^''^ same multiple is the solid LV of the solid AV : For the same 
reason, whatever multiple the base NF is of the base HF, the 
same multiple is the solid NV of the solid ED : And if the base 
LF be equal to the base NF, the solid LV is equal^ to the 
solid N V ; and if the base LF be greater than the base NF, the 
solid LV is greater than the solid NV ; and if less, less : Since 
then there are four magnitudes, viz. the two bases AF, FH, 



^ c. 11. 



* 5. def. 5. 




and the two solids AV, ED, and of the base AF and solid AV, 
the base LF and solid LV are any equimultiples whatever ; and 
of the base FH and solid ED, the base FN and solid N V are 
any equimultiples whatever ; and it has been proved, that if the 
base LF is greater than the base FN, the solid LV is greater 
than the solid NV i and if equal, equal; and if less, less. 
Therefore'' a« the base AF is to the base FH, so is the solid AV 
to the solid ED. Wherefore, if a solid, &c. Q^ E. D. 



PROP. XXVL THEOR. 



SeeN. 



« 11. 11. 



w23. 1. 



'12.11. 



J^T a given point in a given straight line, to make 
a solid angle equal to a given solid angle contained 
by three plane angles. 

Let AB be a given straight line, A a given point in it, and 
D a given solid angle contained by the three plane angles 
EDC, EDF, FDC : It is required to make at the point A in 
the straight line AB a solid angle equal to the solid angle D. 

In the straight lineDF take any point F, from which draw 
"FG perpendicular to the plane EDC, meeting that plane in 
G; join DG, and at the point A, in the straight line AB, 
make'' the angle BAL equal to the angle EDC, and in the 
plane BAL make the angle BAK equal to the angle EDG j 
then make AK equal to DG, and from the point K ercd"' KH 

at 



OF EUCLID. 219 

at right angles to the plane BAL: and make KH equal to GF, Book XI. 
and join AH : Then the solid angle at A, which is contained ^-^^"^^ 
by the three plane angles BAL, BAH, HAL, is equal to the 
solid angle at D contained by the three plane angles EDC, 
EDF, FDC. 

Take the equal straight lines AB, DE, and join HB, KB, 
FE, GE : And because FG is perpendicular to the plane EDC, 
it makes right angles ^ with every straight line meeting it in <3.def. 11., 
that plane : Therefore each of the angles FGD, FGEisa right 
angle: For the same reason, HKA,H1CB are right angles: And 
because KA, AB areequal to GD, DE, each to each, and con- 
tain equal angles, therefore the base BK is equal' to th^ base ' "*• '• 
EG: And KH is equal to GF, and HKB, FGE are right 
angles, therefore HB is equal* to FE: Again, because AK, KH 
arc equal to DG, GF : and contain right angles, the base AH 
is equal to the base DF : and AB is equal to DE : therefore 
HA, AB, are equal to FD, DE, and the base HB is equal to the 
base FE, therefore, 
the angle BAH is J^ 

equal*^ to the angle ^ ^ fg , 

EDF : For the same 
reason, the angle 
HAL is equal to the 
angle FDC. Be- 
cause if AL and DC 
be made equal, and 
KL, HL, GC, FC 

be joined, since the whole angle BAL is equal to the whole 
EDC, and the parts of them'BAK,EDG are, by the construc- 
tion, equal : therefore the remaining angle KAL is equal to 
the remaining angle GDC : And because KA, AL are equal 
to GD, DC, and contain equal angles, the base KL is equal* 
to the base GC : And KH is equal to GF, so that LK, KH are 
equal to CG, GF, and they contain right angles; therefore 
the base HL is equal to the base FC : Again, because HA, AL 
are equal to FD, DC, and thi.- base HL to the base FC, the 
angle HAL is equal' to the angle FDC: Therefore, because 
the three plane angles BAL, BAH, HAL, which contain the 
solid angle at A, are equal to the three plane angles EDC, 
EDF, FDC, v/hich contain the solid angle at D, each to each, 
and are situated in the same order, the s jlid angle at A is equals 
to the solid angle at D. Therefore, at a given point, in a 
jiven straight line, a solid angle has been made equal to a 
given solid angle contained by three plane angles. Which 
was to be done. 





THE ELEMENTS 



^PRdP. XXVII. PROB. 



»26. 11. 



^ 12. 6. 



"24.11. 



X O describe from a 
parallelepiped similar, 
given. 



en straight line k solid 
and similarly situated to one 



Let AB be the given straight line, and CD the given solid 
parallelepiped, .'^t is required from AB to describe a solid 
paralleiopiped similar, and similarly situated to CD. 

At the point A of the given straight line AB, make*, a solid 
angle equal to the solid angle at C, and let BAK, KAH, HAB 
be the three plane angles which contain it, so that BAK be 
equal to the angle ECG, and KAH to GCF, and HAB to 
FCE : And as EC to CG, so make^ BA to AK ; and as GC to 
CF, so make** KA to AH ; wherefore, ex oequali*^, as EC to 

CF, so is BA to AH : Complete the parallelogram BH, and 
the solid AL : And 
because, as EC to 

CG, so BA to AK, 
the sides about the 
equal angles ECG, 
BAK are propor- 
tionals ; therefore 
the parallelogram 
BK is similar to 
EG. For the same 
reason, the paral- 
lelogram KH is similar to GF, and HB to FE. Wherefore 
three parallelograms of the solid AL are similar to three of the 
solid CD; and the three opposite ones in each solid are equal"* 
and similar to these, each to each. Also, because the plane 
angles which contain ths solid angles of the figures are equal, 
each to each, and situated in the same order, the solid angles 

'^H def 11 ^^^ equal=, each to each. Therefore the solid AL is similar*' 
to the solid CD. Wherefore from a given straight line AB a 
solid paralleiopiped AL has been described similar, and simi- 
larly situated to the given one CD. Which was to be done. 




Fz 



«^1 



D 



c 



E 



QF EUCLID, 



PROP. XXVIII. THEOR, 




IF a solid parallelopiped be cut by a plane passing see n. 
through the diagonals of two of the opposite planes; 
it shall be cut in two equal parts. 

Let AB be a solid paraIlelopip,ed, and DE, CF the diago- 
nals of the opposite parallelograms AH, GB, viz, those which 
are drawn betwixt the equal angles in each : And because CD, 
FE are each of them parallel to GA, and not in the same plane 
with itj CD, FE are parallel* ; wherefore the diagonals CF, *^- ^J- 
DE are in the plane in which the pa- 
rallels are, and are themselves paral- 
lels*': And the plane CDEF shall cut 
the solid AB into two equal parts. 

Because the triangle CGF is equaP 
to the triangle CBF, and the triangle 
DAE to DHEj and that the paral- 
lelogram CA is equal"" and similar to 
the opposite one BE j and the paral- 
lelogram GE to CH -, Therefore the 
prism contained by the tv.'o triangles 
CGF, DAE, and the three parallelograms CA, GE, EC, is 
equal'^to the prism contained by the two triangles CBF, DHE, ' C. ii 
and the three parallelograms BE, CH, EC ; because they are 
contained By the same number of equal and similar planes, alike 
situated, and none of their solid angles are contained by more 
than three plane angles. Therefore the solid AB is cut into 
two equal parts by the plane CDEF. Q^ E. D. 

*N. B. The insisting straight lines of a parallelopiped, men- 

* tioned in the next and some following propositions, are the 
^ sides of the parallelograms betwixt the base and the opposite 

* plane parallel to it.' 




16. 11^ 



«^34. 1. 



^2i. 11. 



PROP. XXIX. THEOR. 

oOLID parallelopipeds upon the same base, and of See x. 
the same altitude, the insisting straight lines of 
which are terminated in the same straight lines in the 
plane opposite to the base, are equal to one another. 

Let 



k 



222 
Book XI. 




»28. 11. 



»34. 1. 

'38. 1. 
* 36. 1. 

e24. 11. 



THE ELEMENTS 

Let the solid parallelopipeds AH, AK be upon the same base 
AB, and of the same altitude, and let their insisting straight 
lines AF, AG, LM, LN be terminated in the same straight 
line FN, and CD, CE, BH, BK, be terminated in the same 
straight line DK ; the solid AH is equal to the solid AK, 

First, Let the parallelograms DG, HN, which are opposite 
to the base AB, have a common sideHG : Then, because the 
solid AH is cut by the plane AGHC passing through the diago- 
nals AG, CH of the opposite planes ALGF, CBHD, AH is 
cut into two equal parts* by the plane AGHC : Therefore the 
solid AH is double of the prism 
which is contained betwixt 
the triangles ALG, CBH: For 
the same reason, because the 
solid AK is cut by the plane 
I.GHB through the diagonals 
LG, BH of the opposite planes 
ALNG, CBKH, the solid AK 
is double of the same prism which is contained betwixt the 
triangles ALG, CBH, Therefore the solid AH is equal to 
the solid AK. 

But, let the parallelo<rrams DM, EN, opposite to the base, 
have no common side: Then, because CH, CK are parallelo- 
grams, CB is equaP to each of the opposite sides iJU, EK j 
wherefore DH is equal to EK: A.dd, or take away the common 
part HE ; then DE is tqual to HK : Wherefore also the tri- 
angle CDE is equal^ to the triangle BHK : And the parallelo- 
gram DG is equal* to the parallelogram HN : For the same 
reason, the triangle AFG is equal to the triangle LMN, and 
the parallelogram CF js equal* to the parallelogram BM, and 





CG to 5N : for they arc opposite. Therefore the prism which 
is contained by the two triangles AFG, CDE, and the three 
' C. 11. parallelograms AD, DG, GC is equaU to the prism, contain- 
ed by the two triangles LMN, BHK, and the three parallelo- 
grams BM, MK, KL. If therefore the prism LMN B 11 K be 

taken 



OF EUCLID. 



223 



taken from the solid of which the base is the parallelogram Book XI. 
AB, and in which FDKN is the one opposite to it ; and if '"■^^"^-^ 
from this same solid there be taken the prism AFGCDE, the 
remaining solid, viz. the parallelepiped AH, is equal to the 
remaining parallelepiped AK. Therefore solid parallelo- 
pipeds, &c. Q. E. D, 

PROP. XXX. THEOR. 



OOLID parallelopipeds upon the same base, andSecN. 
of the same altitude, the insisting straight lines of 
which are not terminated in the same straight lines in 
theplaneoppositetothebase, areequaltooneanother. 

Let the parallelopipeds CM, CN, be upon the same base AB, 
and of the same altitude, but their insisting straight lines AF, 
AG, LM, LN, CD, CE, BH, BK, not terminated in the same 
straight lines : The solids, CM, CN are equal to one another. 

Produce FD, MH, and NG, KE, and let them meet one 
another in the points O, P, Q,, R ; and join AO, LP, BQ., 
CR} And because the plane LBHM is parallel to the opposite 



N K 




plane ACDF, and that the plane LBHM is that in which are 
the parallels LB, MFlPQ^, in which also is the figure BLPQ ; 
and the plane ACDF is that in which are the parallels AC, 
FDOR, in which also is the figure CAOR ; therefore the 
figures BLFC^, CAOR are in parallel planes : In like manner, 
because the plane ALNG is parallelto the oppositeplaneCBKE, 
and that the plane ALNG is that iii which are the parallels 

AL, 



224 



THE ELEMENTS 



Book XI. AL, OPGN, in which also is the figure ALPO ; and the plane 
'""^'^'^^ CBKEisthat in which are the parallels CB,RQEK, in which 
also h the figure CBQR j therefore the figures ALPO, CBQil 
are in parallel planes : And the planes ACBL, ORQP arc 
parallel j therefore the solid CP is a parallelepiped : But the 
solid Civl, of which the base is ACBL, to which FDHM is the 
» 2«. 11, opposite parallelogram, is equal* to the solid CP, of which the 




base is the parallelogram ACBL, to which ORQP is the one 
opposite; because they are upon the same base, and their insist- 
ing straight lines AF, AO, CD, CR ; LM, LP, BH, BQ 
are in the same straight lines FR, MQ : And the solid CP is 
equal^ to the solid CN : for they are upon the same base ACBL, 
and their insisting straight lines AO, AG, LP, LN; CR, CE, 
BQ^, BK are in the same straight lines ON, RK : Therefore 
the solid CM is equal to the solid CN. Wherefore solid 
parallelopipeds, &c. Q^ E. D. 



SeeN, 



PROP. XXXL THEOR. 

►^ O LID parallelopipeds, which are upon equal bases, 
and of the same altitude, are equal to one another. 

Let the solid parallelopipeds AE, CF, be upon equal bases 
AB, CD, and be of the same altitude j the solid AE is equal 
to the solid CF. 

First, Let the insisting straight lines be at right angles to the 
bases AB, CD, and let the bases be placed in the same plane, 

and 



OF EUCLID. 



225 



and so as that the sides CL, LB be in a straight lincj there- ^oo^ xi, 
fore the straight line LM, which is at right angles to the plane ^"^^-^^^ 
in which the bases are, in the point L, ^s common* to the two 1 13^ u^ 
solids AE, CF ; let the other insisting lines of the solids b e 
AG, HK, BE ; DF, OP, CN : And first, let the angle ALB b e 
equal to the angle CLD ; then AL, LD ?.;<? in a straight line . » 14. 1. 
Produce OD, HB, and let then: meet in Q,^nd complete the 
solid paralleiopiped LR, the base of which is the parallelogram 
lyQ, and of which LM is one of its insisting straight lines': 
Therefore, because the parallelogram AB is equal to CD, as the 
base AB is to the base LQ, so is ■= the base CD to the same c 7 5^ 
LQ : And becaust: the selid paralleiopiped AR is cut by the 
plane LMEB, which is parallel to the opposite planes AK,DRj 
as the base AB is to the base LQ; so is^ the solid AE to the <" 25. n. 
solid LR : For the same reason, because the solid parallopiped 
CR is cut bv the plane LMFD, which is parallel to the opposite 
planes CP,'BR j as 

the base CD to the P F R 

base LQ, so is the \^K 

solid CF to the solid 

LR : But as the 

base AB to the base 

LQ, so the base CD 

to the base LQ, as 

before was proved ; 

Therefore as the 

solid AEto the solid 

LRj.so is the solid CF to the solid LR j and therefore the 

solid AE is equah to the solid CF. * 9. 5. 

But let the solid parallelopipedsSE, CF be upon equal bases 
SB, CD, and be of the same altitude, and let their insisting 
straight lines be at right angles to the bases ; and place the 
bases SB, CD in the same plane,so that CL, LB be in a straight 
line ; and let the angles SLB, CLD be unequal ; the solid SE 
is also in this case equal to the solid CF : Produce DL, TS 
until they meet in A,and from B draw BH parallel to DA ; and 
let HB, OD produced meet in Q, and complete the solids AE, 
LR: Therefore the solid AE, of which the base is the parallelo- 
gram LE, and AK the one opposite to it, is equal *" to the solid f 29. n. 
SE, of which the base is LE, and to which SX is opposite; 
for they are upon the same base LE, and of the same altitude, 
and their insisting straight lines, viz. LA, LS, BH, BT ; MG, 
MV, EK, EX are in the same straight lines AT, GX : And 

Q because 




126 



THE ELEMENTS 



Book XI. because the parallelogram AB is equals to SB, for they are upon 
tS5^^7r '^^ S^tne base LB, and between the same parallels LB, AT : 
and that the base 
SB is equal to the 
base CD ; therefore 
the base A B is equal 
to the base CD, and 
the angle ALB is 
equal to the angle 
CLD : Therefore, 
by the first case, 
the solid AE is 
equal to the solid 
CF ; but the solid AE is equal to the solid SE, as was demon- 
strated ; therefore the solid SE is equal to the solid CF. 

But, if the insisting straight lines AG, HK, BE,LM; CN, 
RS, DF, OP, be not at right angles to the bases AB, CDj in 
this case likewise the solid AE is equal to the solid CF : From 
the points G, K, E, M j N, S, F, P, draw the straight lines 
Ml. 11. GQ,KT,EV,MX;NY,SZ,FI, PU, perpendicular^ to the 
plane in which are the bases AB, CD ; and let them meet it in 
the points Q, T, V, X ; Y, Z, I, U, and join QT, TV, VX, 
XQ; YZ,ZI,ItJ,UY: Then, because GQKT are at right 




A S 




A H Q 1 



"«. 11. angles to the same plane, they are parallel » to one another : 
And MG, EK are parallels ; therefore the plane MQ, ET, of 
which one passes through MG, GQ, and the other through 
EK, KT, which are parallel to MG, GQ, and not in the same 

"15. II. plane with them, are parallel"* to one another: For the same 
reason, the planes M V,GT are parallel toone another : There- 
fore the solid QE is a parallelepiped : In like manner, it may 
be proved, that the solid YF is a parallelopipcd : But, from 
what has been demonstrated, the solid EQ is equal to the solid 
FY, because they are upon equal bases MK, PS, and of the 
samealtitude,andhavetheirinsistingstraightlinesatrightangles 

3 '• 



OF EUCLID. 



127 



to the bases : And the solid EQ is equal ' to the solid AE ; and ^^^ xi. 
the solid FY to the solid CF; because they are upon the same i^^^^ 
bases and of the same altitude: Therefore the solid AE is equal 11. 
to the solid CF : Wherefore solid parallelepipeds, &c. Q. E. D. 



" PROP. XXXII. THEOR. 

♦Solid parallelopipeds which have tlie same alti- s«eN, 
tude, are to one another as their bases. 

Let AB, CD be solid parallelopipeds of the same altitude : 
They are to one another as their bases j that is, as the base AE 
to the base CF, so is the solid AB to the solid CD. 

To the straight line FG apply the parallelogram FH equal^ » c«r.45 1 
to AE, so that the angle FGH be equal to the angle LCG ; 
and complete the solid parallelepiped GK upon the base FH, 
one of whose insisting lines is FD, whereby the solids CD,GK 
must be of the same altitude : Therefore the solid AB is equal*" * 31. 11. 
to the solid 



B 



N 




GK, because 
they are upon 
equal bases x/ » 

AE, FH, and ^r MO 

are of the same 
altitude: And 

because the so- 

lid parallelopi- A. K 

ped CK is cut 

by the plane DG which is parallel to its opposite planes, the 
base HF is = to the base FC, as the solid HD to the solid DC : c 23. 11. 
But the base HF is equal to the base AE, and the solid GK to 
the solid AB : Therefore, as the base AE to the base CF, so 
is the solid x\B to the solid CD. Wherefore solid parallelo- 
pipeds. Sec. Q. E. D. « 

CoR. From this it is manifest, that prisms upon triangular 
bases, of the same altitude, are to one another as their bases. 

Let the prisms, the bases of which are the triangles AEM, 
CFG, and NBO, PDQ the triangles opposite to them, have 
the same altitude ; and complete the parallelograms, A E,CF, 
and the solid parallelopipeds AB, CD,- in the first of which let 
MO, and in the other let GQ be one of the insisting lines. And 
because the solid parallelopipeds AB, CD have the same alti- 
tude, they are to one another as the base AE is to the base ' 

Q 2 CF ; 




<J 28. 11. 



•24. 11. 



"C. 11. 



M.G, 



THE ELEMENTS 

CF ; wherefore the prisms, which are their halves'*, are to one 
another, as the base AE to' the base CF j that is, as the triangle 
AEM to the triangle CFG. 

PROP. XXXIII. THEOR. 

oIAIILAR solid parallelopipeds are one to another 
in the triplicate ratio of their homologous sides. 

Let AB, CD be similar solid parallelopipeds, and the side 
AE homologous to the side CF : The solid AB has to the solid 
CD, the triplicate ratio of that which AE has to CF. 

Produce AE,GE, HE, and in these produced take EK equal 
to CF, EL equal to FN, and EM equal to FR ; and complete 
the parallelogram KL,and the solid KO : Because KE, EL, are 
equal to CF, FN, and the angle KEL equal to the angle CFN, 
because it is equal to the angle AEG, which is equal to CFN, 
by reason that the solids AB^CD are similar; therefore the 
parallelogram KL is similar and equal to the parallelogram CN. 
For the same reason, the parallelogram MK is similar and equal 
to CR, and also 
OEtoFD. There- 
fore three paralle- < J) 
lograms of the \ N. 

sol id KO are equal 
and similar to three 
parallelograms of \ 
the sol id CD: And 
the three opposite 
ones in each solid 
are equal » and 
similar to these : 

7 herefore the so- O 

fid KO is equal'' 

and similar to the solid CD: Complete the parallelogram 
GK, and complete the solids EX, LP upon the bases GK, 
KL, so that EH be an insisting straight line in each of them, 
whereby they must be of the same altitude with the solid AB : 
And because the solids AB, CD are similar, and, by permu- 
tation, as AE is to CF, so is EG to FN, and so is EH to 
FR ; and B^C is equal to FK, and FN to EL, and FR to EM : 
Therefore, as AP^ to EK, so is EG to EL, and so is HE to 
EM: But, as AE to EK, so"^ is the parallelogram AG to 
the parallelogram GK ; and as GE to EL, so is " GK to KL, 

and 



N 



^R 



c :f 





B 


X 


\ 




\V 


i\ 


P 




G 




\ 


\ 


K 


\ 


\ 




J 


■ML 


\ 


U 




\l 




\ 



OF EUCLID. - 229 

and as HE to EM, 50"= is PE to KM : therefore as the par?.]- ^°^^^- 
lelogram AG to the paralle]o:jram GK, so is GK to KL, and c j g 
PE to KM : But as AG, to GK, so ^ is the solid AB to the «« 25. ii. 
solid EX ; and as GK to KL, so^ is the solid EX to the so- 
lid PL ; and as PE to KM, so-i is the solid PL to the solid KO : 
And therefore as the solid AB to the solid EX, so is EX to 
PL, and PL to KO : But if four magnitudes be continual pro- 
portionals, the first is said to have to the fourth the triplicate 
ratio of that which it has to the second : Therefore the solid 
AB has to the solid KO, the triplicate ratio of that vhich AB 
has to EX : But as AB is to EX, so is the parallelogram AG 
to the parallelogram GK, and the straight line AE to the 
straight line EK. Wherefore the solid AB has to the solid KO, 
the triplicate ratioof that which AE has to EK. And the solid 
KO is equal to the solid CD, and the straight line EK isequal 
to the straight line CF. Therefore the solid AB has to the 
solid CD, the triplicate ratio of that which the side AE has to 
the homologous side CF, kc. Q^ E. D. 

Cor. From this it is manifest, that, if four straight lines be 
continual proportionals, as the first is to the fourth, so is the 
solid parallelopiped described from the first to the similar solid 
similarly described from the second ; because the first straight 
line has to the fourth the triplicate ratio of that which it has 
to the second. 



PROP. D. THEOR. 

OOLID parallelopipeds contained by parallelo- see n. 
grams equiangular to one another, each to each, 
that is, of which the solid angles are equal, each to 
each, have to one another the ratio M'hich is the 
same with the ratio compounded of the ratios of 
their sides. 

Let AB, CD be solid parallelopipeds, of which ABJs con- 
tained by the parallelograms AE, AF, AG equiangular, each 
to each, to the parallelograms CH, CK, CL, which contains 
the solid CD. The ratio which the solid AB has to the solid 
CD, is the same with that which is compounded of the ratios 
of the sides AM to DL, AN to DK, and AO to DH. 

Q_3 Produce 



23® 



Book XI, 



•C. 11. 



*32. 11. 



THE ELEMENTS 

Produce MA, NA, OA to P, Q^, R, so that AP be equal 
to DL, AQ to DK, and AR to DH ; and complete the solid 
parallelopiped AX contained by the parallelograms AS, AT, 
AV similar and equal to CH, CK, CL, each to each. There- 
fore the solid AX is equal* to the solid CD. Complete likewise 
the solid AY, the base of which is AS, and of which AO is 
one of its insisting straight lines. Take any straight line a, 
and as MA to AP, so make a to b ; and as NA to AQj so 
make b to c ; and as AO to AR, so c to d : Then, because the 
parallelogram AE is equiangular to AS, AE is to AS, as the 
straight line a to C, as is demonstrated in the 23d Prop. Book 
6, and the solids AB, AY, being betwixt the parallel planes 
BOY, EAS, are of the same altitude. Therefore the solid AB 
is to the solid AY, as"* the base AE to the base AS : that is, 
as the straight line a is to c. And the solid AY is to the solid 



D 



H 



L 



\ 




\ 


K 






\ 




\ 



a — 
•b — 
c — - 
d— 



C tNl 




AX, as' the base OQ^is to the baseQR ; that is, as the straight 
■^ 25. 11. line OA to AR ; that is, as the straight line c to the straight 
line d. And because the solid AB is to the solid AY, as a is to 
c, and the solid AY to the solid AX, as c is to d } ex acquali, 
the solid AB is to the solid AX or CD which is equal t» 
it, as the straight line a is to d. But the ratio of a to d is said to 
*Def. A. 5. be compounded'^ of the ratios of a to b, b to c, and c to d, 
which are the same with the ratios of the sides MA to AP, 
N A to A(^, and OA to AR, each to each. And the sides AP, 
AQ_, AR are equal to the sides DL, DK, DH, each to each. 
Therefore the solid AB has to the solid CD the ratio, which 
is the same with that which is compounded of the ratios of the 
sides AM to DL, AN to DK, and AQ to DH. Q. E. D. 



OF EUCLID 



231 



Book XI. 



PROP. XXXIV. THEOR. "-^ 

The bases and altitudes of equal solid parallelo- seeN. 
piped, are reciprocally proportional; and if the 
bases and altitudes be reciprocally proportional, the 
solid parallelopipeds are equal. 

Let AB, CD be equal solid parallelopipeds ; their bases are 
reciprocally proportional to their altitudes; that is, as the base 
EH is to the base NP, so is the altitude of the solid CD to the 
altitude of the solid AB. 

First, Let the insisting straight lines AG, EF, LB, HK ; 
CM, NX, OD, PR be at right angles to the bases. As the base 
EH to the base NP, so is 
CM to AG. If the base 
EH be equal to the base 
NP, then because the 
solid AB is likewise equal 
to the solid CD, CM shall 
beaqualtoAG. Because 
if the bases EH, NP, be 
equal, but the altitudes 
AG, CM be not equal, 

neither shall the solid AB be equal to the solid CD. But the 
solids are equal, by the hypothesis. Therefore the altitude CM 
is not unequal to the altitude AG ; that is, they are equal. 
Wherefore, as the base EH to the base NP, so is CM to AG. 

Next, let the bases EH, NP not be equal, but EH greater 
than the other : Since then the solid AB is equal to the solid 
CD, CM is therefore 




greater than AG : For 
if it be not, neither 
also in this case would 
the solids AB, CD be 
equal, which, by the 
hypothesis, are equal. 
Make then CT equal 
to AG, and complete 
the solid parallelopiped 
CV of which the base 
is NP, and altitude 



R_D 



K B 



H 



\ 

G 

\1 




L 



M 



A 



I\x 



-^ 



O 



C N 



CT. Because the solid 

AB is equal to the solid CD, therefore the solid AB is to the 

0^4 solid 



232 



THE ELEMENTS 




Book XI. solid CV, as* the solid CD to the solid CV. But as the solid 
AB to the solid CV, so"* is the base Eri to the D^se NP ; for 
the solids AB, CV are of the same altitude ; and as the solid 
CD to CV, so= is the base MP to the base PT, and s./ is 
the straight line MC to CT ; and CT is crquai to AG. 
Therefore, as the base EH to tht base NP, so is MC to AG. 
Wherefore the bases of the- solid parallelopipeds AB, CD are 
reciprocally proportional to their altitudes. 

Let now the bases of the solid paiallelopipeds AB, CD be 
reciprocally proportional to their altitudes j viz. as the base EH 
to the base NP, so the al- 
titude of the solid CD to 
the altitude of the solid 
AB J the solid AB is 
equal to the solid CD. Let 
the insisting lines be, as 
before, at right angles to H 
the bases. Then, if the 
base EH be equal to the 
base NP, since EH is to 

NP, as the altitude of the solid CD is to the altitude of the 
solid AB, therefore the altitude of CD is equal'^ to the altitude 
of AB. But solid parallelopipeds upen equal bases, and of the 
same altitude, are equal ^ to one another j therefore the solid 
AB is equal to the solid CD. 

But let the bases EH, NP be unequal, and Jet EH be the 
greater of the two. Therefore, since as the base EH to the base; 
NP, so is CM thealti- 



K 


B 


1 


PL 


V 


► 


\ 


(r \ 


"5? 


\M 


\- 






T, 


P 









\ 




\ 




\ 




\ 


1 


\ 


t 


( 


* 


N 



sA. 5. 



'31. 11. 



tude of the solid CD to 
AG the al titude of AB, 
CM is greater'' than 
AG. Again, take CT 
equal to AG, and com- 
plete, as before, the so 
lid CV. And because 
the base EH is to the 
baseNP,asCMtoAG 
and that AG is equal to 
CT, therefore the base 



R D 



K I^ 



H 






\ 




\ 


M 




K 


N 

T 









\ 




\ 



A E 



C N 



X 



EH is to the base NP, as MC to CT. But as the base EH 
is to NP, so^ is the solid AB to the solid CV ; for the solids 
A3> CV are of the same altitude j and as MC to CT, so i$ 

the 



OF EUCLID. 



233 



the base MP to the base PT, and the solid CD to the solid' ^^^V^J* 
CV : And therefore as the solid AB to the solid CV, so is the c s^^^nT 
solid CD to the solid CV ; that is, each of the solids AB, CD 
has the same ratio to the solid C V : and therefore tha solid AB 
is equal to the solid CD. 

Second general case. Let the insisting straight lines FE, 
BL, GA, Kf^ ; XN, DO, iVlC, RP not be at right angles to 
the bases of rhe solids ; and from the points F, B, K,'G ; X, 
D, R, M draw perpendiculars to the planes in which are the 
bases EH, NP meeting those planes in the points S, Y, V, T; 
Q, 1, U, Z ; and complete the solids FV, XU, which are 
parallelnpipeds, as was proved in the last part of Prop. 31 of 
this Book. In this case, lilcewise, if the solids AB, CD be 
equal, their bases are reciprocally proportional to their altitudes, 
viz. the base EH to the base NP, as the altitude of the solid 
CD to the altitude of the solid AB. Because the solid AB is 
equal to the solid CD, and that the solid BT is equal? to the e'29or30. 
solid BA, for they are upon the same base FK, and of the ^^' 





same altitude; and that the solid DC is equals to the solid 
DZ, being upon the same base XR, and of the same altitude ; 
therefore the solid BT is equal to the solid DZ : But the bases 
are reciprocally proportional to the c 'titudes of equal solid 
parallelepipeds of r/hich the insisting straight lines are at right 
angles to their bases, as before was proved : Therefore as the 
base FK to the base XR, so is the altitude of the solid DZ to 
the altitude of the solid BT : And the base FK is equal to the 
base EH, and the base XR to the base NP : Wherefore, as the 
base EH to the base NP, so is the altitude of the solid DZ to 
the altitude of the solid BT : But the altitudes of the solids 
DZ, DC, as also of the solids BT, BA are the same. There, 
fore as the b^se EH to the br;se NP, so is the altitude of the 

solid 



»34 



THE ELEMENTS 



Book XI. solid CD to the altitude of the solid AB ; that is, the bases of 
'"^^'^^'^^ the solid parallelopipeds AB, CD are reciprocally proportional 
to their altitudes. 

Next, Let the bases of the solids AB, CD be reciprocally 
proportional to their altitudes, viz. the base EH to the base 
NP, as the altitude of the solid CD to the altitude of the solid 
ABj the solid AB is equal to the solid CD: The sanae 
construction being made ; because, as the base EH to the base 
NP, so is the altitude of the solid CD to the altitude of the 
solid AB ; and that the base EH is equal to the base FK; and 
NP to XR ; therefore the base FK is to the base XR, as the 
altitude of the solid CD to the altitude of AB : But the alti- 





tudes of the solids AB, BT are the same, as also of CD and 
DZ i therefore as the base FK to the base XR, so is the alti- 
tude of the solid DZ to the altitude of the solid BT : Where- 
fore the bases of the solids BT, DZ are reciprocally propor- 
tional to their altitudes: and their insisting straight lines are 
at right angles to the bases ; wherefore, as was before proved, 
• 29 or 30. the solid BT is equal to the solid DZ; But BT is equals to 
"• the solid BA, and DZ to the solid DC, because they are upon 

the same bases, and of the same altitude. Therefore the solid 
AB is equal to the solid CD. Q^ E. D. 



OF EUCLID. 
PROP. XXXV. THEOR. 



235 

Book XI. 



XF, from the vertices of two equal plane angles, there See n. 
be drawn two straight lines elevated above the 
planes in which the angles are, and containing equal 
angles with the sides of those angles, each to each ; 
and if in the lines above the planes there be taken 
any points, and from them perpendiculars be drawn 
to the planes in which the first named angles are : 
And from the points in which they meet the planes, 
straight lines be drawn to the vertices of the angles 
first named ; these straight lines shall contain equal 
angles with the straight lines whicli are above the 
planes of the angles. 

Let BAC, EDF be two equal plane angles ; and from the 
points A, D let the straight lines AG, DM be elevated above 
the planes of the angles making equal angles with their sides, 
each to each, viz. the angle GAB equal to the angle MDE, and 
GAC to MDF ; and in AG, DM let any points G, M be 
taken, and from them let perpendiculars GL, MN be drawn to 
theplanes BAC, EDF, meeting these planes in the pointy L, Nj 

U 




and join LA, ND : The angle' GAL is equal to the angle 

Make AH equal to DM, and through H draw HK parallel 
to GL: But GL is perpendicular to the plane BAC; where- 
fore HK is perpendicular" to the same plane : From the points 
K, N, to the straight lines AB, AC, DE, DF, draw perpen- 
diculars KB, KC, N£, NFj and join HB, BC, ME, EF : 

Because 



8. 11. 




«4.clef. 11, 
'S.def.ll, 



e 26. 1. 



THE ELEMENTS 

Because HK is perpendicular to the plane BAC, the plane 
HBK v/hich passes through HK is at rightangles*' to the plane 
BAC ; and AB is drawn in the plane BAC at right angles to 
the con^mon section BK of the two planes ; therefore AB is 
perpendicular'^ to the plane HBK, and makes right angles'" 
with every straight line meetingitin that plane: But BH meets 
it in that plane ; therefore ABH is a right angle : For the same 
reason, DEM is a right angle, and is therefore equal to the 
angle ABH : And the angle HAB is equal to the angle MDE. 
Therefore in the two triangles HAB, MDE there are two 
angles in one equal totwoangles in the other, each to each, and 
one side equal to one side, opposite to one of the equal angles 
in each, viz. HA equal to DM j therefore the remaining sides 
are equal<=, each to each : Wherefore AB is equal to DE. In 
the same manner, if HC and MF be joined, it may be demon- 
strated that AC is equal to DF : Therefore, since AB is equal 
to DE, BA and AC are equal to ED and DF j and the angle 




BAC is equal to the anle EDF j wherefore the base BC 
■4. i; equaF to the base EF, and the remaining angles to the remain- 
ing angles : The angle ABC is therefore equal to the angle 
DEF : And the right angle ABK is equal to the right angle. 
DEN, whence the remaining angle CBK is equal to the 
remaining angle FEN : For the same reason, the angle BCK is 
equal to the angle EFN : Therefore in the two triangles BCK, 
EFN, there are two angles in one equal to two angles in the 
other, each to each, and one side equal to one side adjacent 
to the equal angles in each, viz. BC equal to EF ; the other 
sides, therefore, are equal to the other sides ; BK then is equal 
to EN: and AB is equal to DE ; wherefore AB, BK are equal 
to DE, EN ; and they contain right angles ; wherefore the 
base AK is equal to the br.sc DN : And since AH is equal to 



OF EUCLID. 23f 

DM, the square of AH Is equal to the square of DM: But the Book xr. 
squares of AK, KH are equal to the squares of AH, because g^^i 
AKH is a right angle : And the squares of DN, NM are equal 
to the square of DM, for DNM is a right angle: Wherefore 
the iquarcs of AK; KH are equal to the squares of DN, NM; 
and of those the square of AK is equal to the square of DN : 
Therefore the remainincr square of KH is equal to the remain- 
ing square of NM ; and the straight line KH to the straight 
line NM : and because HA, AK are equal to MD, DN, each 
to each, and the base HK to the base MN, as has been proved ; 
therefore the angle HAK is equal ^ to die angle MDN. '"S, i. 
Q, E. D. 

Cor. From this it is manifest, that if, from the vertices of 
two equal plane angles, there be elevated two equal straight 
lines containing equal angles with the sides of the angles, each 
to each j the perpendiculars drawn from the extremities of the 
equal straight lines to the planes of the first angles are equal 
to one another. 



Another Demonstration of the Corollary. 

Let the plane angks BAC, EDF be equal to one another, 
and let AH, DM, be two equal straight lines above the planes 
of the angles, containing equal angles with BA, AC ; ED 
DF, each to each, viz. the angle HAB, equal to MDE, and 
HAC equal to the angle MDF ; and from H, M let HK, MN 
be perpendiculars to the planes BAC, EDF : HK is equal to 
MN.. 

Because the solid angle at A is contained by the three plane 
angles BAC, BAH, HAC, which are, each to each, equal to 
the three plane angles EDF, EDM, MDF containing the solid 
angle at D ; the solid angles at A and D are equal, and there- 
fore coincide withone another ; to wit, ifthe plane angle BAC 
be applied to the plane angle EDF, the straight line AH coin- 
cides with DM, as was shewn in Prob. B. ol" this Book : And 
because AH is equal to DM, the point H coincides with the 
point M: Wherefore HK, which is perpendicular to the plane 
BAC, coincides with MN» which is perpendicular to the plane ' 13. u. 
EDF, because these planes coincides v/ith one another : 
Therefore HK is equal to MN. Q. E. D. • 



238 



THE ELEMENTS 



Book XI. 



SteN. 



•26.11. 



14.6. 



PROP. XXXVI. THEOR. 

1 F three straight lines be proportionals, the solid 
parallelopiped described from all three as its sides, 
is equal to the equilateral parallelopiped described 
from the mean proportional, one of the solid angles 1 
of which is contained by three plane angles equal, \ 
each to each, to the three plane angles containing 
one of the solid angles of the other figure. 

Let A, B, C be three proportionals, viz. A to B, as B to C: . 
The solid described from A, B, C is equal to the equilateral 
solid described from B, equiangular to the other. ; 

Take a solid angle D contained by three plane angles EDF, 
FDG, GDE; and make each of the straight lines ED, DF, ' 
DG equal to B, and complete the solid parallelopiped DH : j 
Make LK equal to A, and at the point K in the straight linef 
LK, make ^ a solid angle contained by the three plane angles 
LKM, MKN, NKL equal to the angles EDF, FDG, GDE, 



H 










\ 


' 


\ 


\ 




X 






M 




N 






F 


\ 




\ 


N 




\ 



G 



K 



E 



D 



B 



each to each; and make KN equal to B, and KM equal to 
C; and eomplete the solid parallelopiped KG : And because, as 
A is to B, so is B to C, and that A is equal to LK, and B 
to each of the straight lines DE, DF, and C to KM ; there- 
fore LK is to ED, as DF to KM ; that is, the sides about the 
equal angles are reciprocally proportional j therefore the pa- 
rallelogram LM is equal^ to EF; and because EDF, LKM 
are two equal plane angles, and the two cciual straight lines DG, 
KN are drawn from their vertices above their planes, and con- 
tain equal angles with their sides j therefore the perpendi- 
culars from the points G, N, to the planes EDF, LKM are 

equal 



OF EUCLID. 



239 



equal*^ to one another : Therefore the solids KO, DH are of ^^^J^^ 
the same altitude ; and they are upon equal bases LM, EF, c Co^ 33. 
and therefore they are equal'' to one another: But the solid n. 
KO is described from the three straight lines A, B, C, and *^^- *'* 
the solid DH from the straight line B. If therefore three 
straight lines, &c. Q^ E. D. 

PROP. XXXVII. THEOR. 

Xf four straight lines be proportionals, the similar s«3f- 
solid parallelopipeds similarly described from them 
shall also be proportionals. And if the similar pa- 
rallelopipeds similarly described from four straight 
lines be proportionals the straight lines shall be 
proportionals. 

Let the four straight lines AB, CD^ EF, GH be propor- 
tionals, viz. as AB to CD, so V.F to GH ; and let the simi- 
lar parallelopipeds AK, CL, EM, GN be similarly described 
from them. AK is to CL, as EM to GN. 

Make* AB, CD, O, P continual proportionals, as also EF, »n.6. 
GH, Q^ R : And because as AB is to CD, so EF to GH : and 



K 



B 



C2J 












3 



S 



d 



K 



that CD is*" to O, as GH toQ_, and O to P, as Q,, to R ; there- » n. 5. 
fore, ex aequali'', AB is to P, as EF to R : But as AB to P, c22. 5, 
SO"* is the solid AK, to the solid CL ; and as EF to R, so'' is " ^'"'■•33. 
the solid EM to the solid GN : Therefore'' as the solid AK to 
the solid CL, so is the solid EM to the solid GN. 

But 



11. 



i4^ 



THE ELEMENTS 



*2Y. n. 



Book Xf. But let the solid AK be to the solid CL^ as the solid EM to 
the solid GN : The straight line AB ib to CD, as EFto GH. 
Take AB to CD, as Ev to ST, and from ST describe <= a 
solid parallelepiped SV similar and similarly situated to either 
of the solids £M, GN : And because AB is to CD, is EF to 
ST, and that from AB, CD the solid parallelopipeds AK, CL 
are similarly described j and in like manner the solids EM, SV 
from the straight lines EF, ST 3 therefore AK is to CL, as 



i. 



*9.5. 



SccN, 





Kl 


^ 


N 


1 


\ 


1 \ 


■ J 


I B 



K 








1) 




(; H 



d 



R 



EM to SV : But, by the hypothesis, AK is to CL, as EM to 
GN : Therefore GN is equaK to SV : But it is like'A^ise similar 
and similarly situated to S V; therefore the planes which contain 
the solids GN, SV are similar and equal, and their hbmologous 
sides GH, ST equal to one another : iVnd because as AB to CD, 
so EF to ST, and that ST is equal to GH: AB is to CD, as 
EF to GH. Therefore, if four straight lines, &c, Q. E. D, 

PROP. XXXVHL THEOR. 

If a plane be perpendicular to another plane, and 
" a straight line be drawn from a point in one of 
" the planes perpendicular to the other plane, this 
*' straight line shall fall on the common section of 
" the planes." 

" Let the plane CD be perpendicular to the plane AB, and 
*' let AD be the common se«Sion j if any point E be taken in 
" the piane CD, the perpendicular drawn from E to the plane 
« AB shall fall on AD. 

3 " f Of 



OF EUCLID. 



141 



«3,def.lL 



« For, if it does not, let it, if possible, fall elsewhere, as EF ; ^^^;^- 
" and let it meet the plane AB in the point F j and from F 
« draw% in the plane AB a perpendicular FG to DA, which » 12. 1. 
•' is also perpendicular *" to the plane CD j and join EG : Then » 4. dcf. 11. 
" because FG is perpendicular 
•* to the plane CD, and the 
** straight line EG, which is in 
•* that plane, meets it j there- 
** fore FGE is a right angle'^ : 
** But EF is also at right angles 
** to the plane AB ; and there- 
*' fore EFG is a right angle : 
** Wherefore two of the angles 

" of the triangle EFG are equal together to two right angles ; 
*' which is absurd : Therefore the perpendicular from the 
" point E to the plane AB, does not fall elsewhere than upon 
" the straight line AD; it therefore falls upon it. If there- 
« fore a plane," &c. Q^ E. D. 




PROP. XXXIX. THEOR. 

In a solid parallelopiped, if the sides of two of the SeeN. 
opposite planes be divided, each into two equal parts, 
the common section of the planes passing through 
the points of division, and the diameter of the solid 
parallelopiped cut each other into two equal parts. 

Let the sides of 

the opposite planes 11 K K 

CF, AH of the 
solid parallelopiped 
AF, be divided 
each into two 
equal parts in the 
points K, L, M, 
N; X, O, P, R; 
and join KL, MN, 
XO, PR; And be- 
cause DK, CL are 
equal and parallel, 

KL is parallel* to ^ H^^^-X I |^* XI "33. li. 

DC : For the same 
reason, MN is pa- 
rallel to BA : And 




•242 
Book. XI. 

»5. 11. 



« 29. 1. 



« 4. 1. 



«U. 1. 



"33, 1. 



* 15. 1. 



8 26. 1. 



THE ELEMENTS 

BA is parallel to DC j therefore, because KL, BA are each of 
them parallel to DC, and not in the same plane with it, KL is 
parallel'' to BA : And because KL, MN are each of them 
parallel'to BA, and not in the same plane with it, KL is parallel 
^'to MN; wherefore KL, MN are in one plane. In like 
manner it may be proved, that XO, PR arte in one plane. Let 
YS be the common section of the planes KN, XR ; and DQ 
the diameter of the solid parailelopiped AF : YS and DG dd 
ipeet, and cut one another into two equal parts. 

Join DY, YE, BS, SG. Because DX is parallel to OE, the 
alternate angles DXY, YQE are equal"^ to one another : And 
because DX is 
equal to OE, and 
XY to YO, and 
contain equal an- 
gles, the base DY 
is equal ^ to the 
base YE, and the 
other angles are 
equal ; therefore 
the angle XYD 
is equal to the 
angle OYE, and 
Dye is a straight" 
* line : For the 
same reason BSG 
is a straight line, 
and BS equal to 
SG : And because CA is equal and parallel to DB, and also 
equal and parallel to EG; therefore DB is equal and parallel^ 
to EG : And DE, BG join their extremities ; therefore DE is 
equal and parallel * to BG : And DG, YS are drawn from 
po;nts in the one, to points in the other ; and are therefore in 
one plane : Whence it is manifest, that DG, YS must mtet 
one another ; let them meet in T : And because DE is pa- 
rallel to BG, the alternate angles EDT, BG I' are equal*^: 
and the angle DTY is equaK to the angle GTS : Therefore 
in the triangles DTY, GTS there are two angles in the one 
equal to two angles in the other, and one side equal to one side, 
opposite to two of the equal angles, viz. DY to GS ; for they 
are the halves ot DE, BG : Therefore the remaining sides are 
equaiE, each to each. Wherefore DT is equal to TG, aod 
YT equal to TS. Wherefore, if in a solid, 6ic. Q. E. D. 




OF EUCLID. 



PROP. XL. THEOR. 

XF there be two triangular prisms of the sanie al- 
titude, the base of one of which is a parallelogram, 
and the base of the other a triangle ; if the paral- 
lelogram be double of th« triangle, the prisms shall 
be equal to one another. 

Let the prisms ABCDEF, GHKLMN be of the same alti- 
tude, the first whereof is contained by the two triangles ABE, 
CDF, and the three parallelograms AD, DE, EC ; and the 
other by the two triangles GHK, LMN, and the three paral- 
lelograms LH, HN, NG J and let one of them have a paral- 
lelogram AF, and the other a triangle GHK for its base ; if 
the parallelogram AF be double of the triangle GHK, the 
prism ABCDEF is equal to the prism GHKLMN. 

Complete the solids AX, GO ; and because the parallelo- 
gram AF is double of the triangle GHK j and the parallelo- 





gram HK double* of the same triangle; therefore the paral- »34. i. 
lelogram AF is equal to HK. But solid parallelopipeds upon 
equal bases, and of the same altitude, are equaP to one another, b ^j^ jj 
Therefore the solid AX is equal to the solid GO j and the 
prism ABCDEF is halt* of the solid AX; and the prism <«« ,, 
GHKLMN half ^ of the solid GO. Therefore the prism * 
ABCDEF is equal to the prism GHKLMN. Wherefore, if 
there is two, &c. Q^ E. D. 



R ? 



[ 244 ] 



THE 



ELEMENTS 



or 



EUCLID. 



Book XII. 



BOOK XII. 



SeeN. 



LEMMA I. 

Which is the first preposition of the tenth book, and is neces- 
sary to some of the propositions of this book. 

J.F from the greater of two unequal magnitudes, 
there be taken more than its half, and from the re- 
mainder more than its half; and so on : There shall 
at length remain a magnitude less than the least of 
the proposed magnitudes. 

Let AB and C be two unequal magnitudes, of which AB is 
the greater. If from AB there be taken more 
that its half, and from the remainder more 
than its half, and so on ; there shall at length 
remain a magnitude less than C. 

For C may be multiplied so as at length to 
become greater than AB. Let it be so multi- 
plied, and let DE its multiple be greater than 
AB, and let DE be divided into DF, FG, GE, H 
each equal to C. From AB take BH greater p 

than its half, and from the remainder AH 
take HK greater than its half, and so on, until 
there be as many divisions in AB as there are 
in DE : And let the divisions in AB be AK, 
KH, HB ; and the divisions in ED be DF, FG, 
GE. And because DE is greater than AB, and 



A 



D 



K 



F 



\ 



B C E 

that 



THE ELEMENTS OF EUCLID. 



245 



that EG taken from DE is not greater than its half, but BH Book xii. 
taken from AB is greater than its half; therefore the remain- ^'^^''^'^ 
der GD is greater than the remainder HA. Again, because 
GD is greater than HA, and that GF is not greater than the 
half of GD, but HK is greater than the half of HA ; therefore 
the remainder FD is greater than the remainder AK. And 
FD is equal to C, tlierefore C is greater than AK j that is, 
AK is less than C. Q. E. D. 

And if only the halves be taken away, the same thing may 
in the same way be demonstrated. 



PROP. I. THEOR. 



oliVIILAR polygons inscribed in circles, arc to 
one another as the squares of their diameters. 

Let ABODE, FGHKL be two circles, and in them the 
similar polygons ABODE, FGHKL j and let BM, GN be the 
diameters of the circles: As the square of BM is to the square 
ofGN, so is the polygon ABODE to the polygon FGHKL. 

Join BE, AM, GL, F\ : And because the polygon ABODE 
is similar to the polygon FGHKL, and similar polygons are di- 
vided into similar triangles ; the triangles ABE , FGL, are similar 





and equiangular'' ; and therefore the angle AEB is equal to the * 6. 6. 
angle FLG: But AEB is equal"^ to AMB, because they stand *2i. 3. 
upon the same circumference ; and the angle FLG is, for the 
same reason, equal to the angle FNG : Therefore also the 
angle AMB is equal to FNG : And the right angle BAM is 
equal to the right** angle GFN ; wherefore the remaining an- * 31. 3, 
gjes in the triangles ABM, FGN are equal, and they are equi- 

R 3 angular 



24^ 



THE ELEMENTS 



Book XII. angular to one another : Therefore as BM to GN,' so* is BA 
^"^o"^"^ to GF ; and therefore the duplicate ratio of BM to GN, is 
f 10 def. 5. the same*^ with the duplicate ratio of BA, to GF : But the ratio 
«ML 6*' °^ ^^^ square of BM to the square of GN, is the duplicates 
ratio of that which BM has to GN j and the ratio of the po- 





lygon ABCDE to the polygon FGHKL is the duplicates of 
that which BA has to GF : Therefore as the square of BM 
to the square of GN, so is the polygon ABCDE to the poly- 
gon FGHKL. Wkerefore similar polygons, &c. Q. E. D. 



PROP. II. THEOR. 



Sfe N. 



V^IRCLES are to one another as the squares of 
their diameters. 



Let ABCD, EFGH be two circles, and BD, FH their dia- 
meters ; As the square of BD to the square of FH, so is the 
circle ABCD, to the circle EFGH. 

For, if it be not so, the square of BD shall be to the square 
of FH, as the circle ABCD is to some space either less than 
the circle EFGH, or greater than it*. First let it be to a space 
S less than the circle EFGH; and in the circle EFGH 
describe the square EFGH. This square is greater than 
half of the circle EFGH ; because if, through the points 
E, F, G, H, there be drawn tangents to the circle, the 

square 

• For tliere is some square equal to the rirck ABCD; let P be the wde of it, and 
to three straight lines BD, FH,and P, tlicre can l>ea fourlh jiroportional ; let this be 
Q : Therelbrc the squares of these four siraighl liiic« are proportionals; that is, to the 
»|uaresof BD, FH, and^the circle ABCB), it is possible there may be a fourth pro- 
portionaL Let this be S. Aud in like inaimcr are to be understood soanc things ii^ 
ioni« of the foUswing proposition*. 



OF EUCLID. 24-7 

square EFGH.is half* of the square described about thecirclci BookXII. 
and the circle is less than the square described about it; there- .^^j 
fore the square EFGH is greater than half of the circle. Divide 
the circumferences EF, FG, GH, HE, each into two equal 
parts in the points, KjLjM,N,arui join EK,KF, FL,T..^, GM, 
MH, HX, NE : Therefore each of the triangles EKF, FLG, 
GMH, HNE is greater than half of the segment of the circle 
it stands in ; because, if straight lines touching the circle be. 
drawn through the points K, L, >I, N, and the parallelograms 
upon the straight lines EF, FG, GH, KE, b: completed ; each 
of the triangles EKF, FLG, GMH, HNE shall be the half^ 
of the parallslogram in which it is ; But every segment is less 
than the parallelogram in which it is : Wherefore each of the 
triangles EKF, FLG, GMH, HNE is greater than half the 
segment of the circle which contains it : And if these circum- 
ferences before named be divided each into two equal parts, 
and their extremities be joined by straight lines, by continuing 




to do this, there will at length remain segments of the circle, 
which, together, shall beless than the excessof thecircle EFGH, 
above the space S : Because, by the preceding Lemma, if 
from the greater of two unequal magnitudes there be taken 
more than its half, and from the remainder more than its 
half, and so on, there shall at length remain a magnitude less 
than the least of the proposed magnitudes. Let then the seg- 
ments EK, KF, FL, LG, GM, MH, HN, NE be those that 
remain, and are together less than the excess of the circle EFGH 
above S: Therefore the rest of the circle, viz. the polygon 
EKFLGMHN, is greater than the space S. Describe likevi'ise 
in the circle ABCD the polygon AXBOCPDR similar to the 
polygon EKFLGMHN : As therefore, the square of BD is to 
the square of VH^, so is the polygon AXBOCPDR co the i> I. '«?. 
polygon EKFLGMHN : But the square ofBD is also to the 

R 4 square 



248 



THE ELEMENTS 



Book XII. square of FH, as the circle ABCD is to the space S . There- 
j^JpC*"^ fore as the circle ABCD is to the space S, so is'= the polygon 
' * AXBOCPDR to the polygon EKFLGMHN : But the circle 
ABCD is greater than the polygon contained in it ; wherefore 
* U.5, the space S isgreater'' than the polygon EKFLGMHN : But 
it is likewise less, as has been demonstrated; which is impos- 
sible. Therefore the square of BD is not to the square of FH, 
as the circle ABCD is to any space less than the circle EFGH; 
In the same manner, it may be demonstrated, that neither is the 
square of FH to the square of BD, as the circle EFGH is t» 
any space less than the circle ABCD. Nor is the square of 
BD to the square of FH, as the circle ABCD is to any space 
greater than the circle EFGH : For, if possible, let it be so to 
T,aspacegreaterthanthecircleEFGH: Therefore, inversely, 
as the square of FH to the square of BDt so is the space T to 




the circle ABCD. But as the spacef T is to the circle ABCD, 
so is the circle EFGH to some space, which must be less*" thaa 
the circle ABCD, because the space T is greater, by hypothe- 
sis, than the circle EFGH. Thereforeas thesquareofFH is to 

the 



+ For, as in the foreeoing note at», it was explained how it was possible there 
could be a fourth proportional to the squares ot BD, FH, and the c.rcle ABCD, 
which was named S. So, in like manner, <^'"f,f»" ^'i! » J""'-'.*' fX^'"""' ^ 
this other space, named T, and the c.rcles ABCD, EFGH. And the hkc » to 
W understood ia some of the following pcooositioni. 



OF EUCLID. 



249 



the square of BD, so is the circle EFGH to a space less than ^^^^ '^^^' 
the circle ABCD, which has been demonstrated to be impos- "'^'^'^^'^ 
sible : Therefore the square of BD is not to the square of FH 
as the circle ABCD is to any space greater than the circle 
EFGH : And it has been demonstrated, that neither is the 
square of BD to the square of FH, as the circle ABCD to any 
space less than the circle EFGH : Wherefore, as the square of 
BD to the square of FH, so is the circle ABCD to the circle 
EFGHf . Circles therefore are, &c. Q^ E. D, 



PROP. III. THEOR. 

ti VER Y pyramid having a triangular base, may be see x. 
divided into two equal and similar pyramids having 
triangular bases, and which are similar to the whole 
pyramid ; and into two equal prisms wliich together 
are greater than half of the whole pyramid. 

Let there be a pyramid of which the base is the triangle ABC 
and its vertex the point D: The pyramid ABCD may be di- 
vided into two equal and similar pyra- 
mids having triangular bases, and simi- 
lar to the whole ; and into two equal 
prisms which together are greater than 
half of the whole pyramid. 

Divide AB, BC,'CA, AD, DB, DC, 
each into two equal parts in the; points 
E, F, G, H, K, L, and join EH, EG, 
GH, HK, KL, LH, KK, KF, FG. 
Because AE is equal to EB, and AH to 
HD, HE is parallel =• to DB : For the 
same reason, HK is parallel to AB; 
Therefore HEBK is a parallelogram, 
and HK equal ^ to EB: but EB is equal 
to AE ; therefore also AE is equal to 
HK : And AH is equal to HD; where- 
fore EA, AH are equal to KH, HD, 
each to each ; and the angle t AH is equal"^ to the angle KHD ; 
therefore the base EH is equal to the baseKD, and the triangle 

AEH 




" 34. 1. 



<=29. 



+ ^fc»"se *»a fourth proportional I0 the square? cf BD, FH, and thf circl- 
ABC D, is possible, atwi that ilcaii-neitLcr be i^» iior greater than the circie iFGK, 
mast \m equal to it. 



250 



10. n. 



<"C. 11. 



«4. 6. 



"21. 6. 



»B. 11. & 
ll.dcf. 11 



"41. 1. 



THE ELEMENTS 

AEH equal^ and similar to the triangle HKD : For the same 
reason, the triangle AGH is equal and similar to the triangle 
HLD : And because the two straight lines EH, HG, which 
meet one another, are parallel to KD, DL that meet one ano- 
ther, and arc not in the same plane WJth them, they contain 
equai^ angles; therefore the angle EHG is equal to the angle 
KDL. Again, because EH, HG are equal to KD, DL, each 
to each, and the angle EHG equal to the angle KDL ; there- 
fore the base EG is equal to the base KL : And the triangle 
EHG equaF and simi,lar to the triangle KDL : For the same 
reason, the triangle AEG is also equal and similar to the tri- 
angle HKL. Therefore the pyramid, of which the base is the 
triangle AEG, and of which the vertex is the point H, is 
equal ^ and similar to the pyramid the 
base of which is the triangle KHL, and 
vertex the point D : And because HK 
is parallel to AB, a side of the triangle 
ADB, the triangle ADB is equiangu- 
lar to the triangle HDK, and their 
sides are proportionals^ ; Therefore the 
triangle ADB is similar to the triangle 
HDK: And for the same reason, the 
triangle DBC is similar to the triangle 
DKL; and the "triangle ADC to the 
triangle HDL ; and also the triangle 
ABC to the triangle AEG: But the 
triangle AEG is similar to the triangle 
HKL, as before was proved ; therefore 
the triangle ABC is similar '^ to the 
triangle HKL. And the pyramid of 
which the base is the triangle ABC, and vertex the point D 
, is therefore S'imilar^ to the pyranjid of which the base is the tri- 
angle HKL, and vertex the same point D: But the pyramid of 
which the base is the triangle HKL, and vertex the point D, is 
similar, as has be n proved, to thepyramid the base of which is 
the triangle AEG, and vertex the point H : Wherefore the 
pyramid, the base of which is the triangle ABC, and vertex the 
point D, is similar to the pyramid of which the base is the tri- 
angle AEG and vertex H : Therefore each of the pyramids 
AEGH, HKLD is similar to the whole pyramid ABCD: And 
because BE is equal to EC, the parallelogram EBFG is double"^ 
of the triangle GFC : But when there are two prisms of the 




OF EUCLID. 251 

same altitude, of which one has a parallelogram for its base, ^'^^ ^^• 
and the other a triangle that is half of the parailelogram, these 
prisms are equal^ to one another ; therefore the prism having ,^ 2jj. 
the parallelogram EBFG for its base, and the straight line 
KH opposite to it, is equal to the prism having the triangle 
GFC for its base, and the triangle HKL opposite to it; for 
they are of the same altitude, because they are between the 
parallel^ planes ABC, HKL : And it is manifest that each of * ^^- ''• 
these prisms is greater than either of the pyramids of which 
the triangles AEG, HKL are the bases, and the vertices the 
points H, D J because, if EF be joined, the prism having the 
parellelogram EBFG for its base, and KH the straight line 
opposite to it, is greater than the pyramid of which the base 
is the triangle EBF, and vertex the point K ; but this pyra- 
mid is equal*^ to the pyramid the base of which is the triangle '^•^*- 
AEG, and vertex the point H ; because they are contained 
by equal and similar planes : Wherefore the prism having the 
parallelogram EBFG for its base, and opposite side KH, is 
greater than the pyramid of which the base is the triangle 
AEG, and vertex the point K : And the prism of which the 
base is the parallelogram EBFG, and opposite side KH is 
equal to the prism having the triangle GFC for its base, and 
HKL the triangle opposite to it; and the pyramid of which 
the base is the triangle AEG, and vertex H, is equal to the 
pyramid of which the base is the triangle HKL, and vertex D : 
Therefore the two prisms before-mentioned are greater than 
the two pyramids of which the bases are the triangles AEG, 
L, and vertices the points H, D. Therefore the whole 
P^ amid of which the base is the triangle ABC; and vertex- 
the point D, is divided into two equal pyramids similar to one 
another, and to the whole pyramid ; and into two equal prisms ; 
and the two prisms are together greater than half of the whole 
pyramid. CL E. D. 



252 THE ELEMENTS 

Book Xir. 

'"'^'^ PROP. IV. THEOR. 

secN. J^p there be two pyramids of the same altitude, 
upon triangular bases, and each of them be divided 
into two equal pyramids similar to the whole pyra-' 
mid, and also into two equal prisms ; and if each of 
these pyramids be divided in the same manner as 
the first two, and so on: As the base of one of the 
first two pyramids is to the base of the other, so 
shall all the prisms in one of them be to all the 
prisms in the other that are produced by the same 
number of divisions. 

Let there be two pyramids of the same altitude upon the tri- 
angular bases ABC, DEF, and having their vertices in the 
points G, H; and let each ot them be divided into two equal 
pyramids similar to the vfhole, rnd into two equal prisms; and 
let each of the pyramids thus made be conceived to be divided 
in the like manner, and so on: As the base ABC is to the base 
DEF, so are all the prisms in the pyramid ABCG to all the 
prisms in the pyramid DEFH made by the same number of 
divisions. 

Make the same construction as in the foregoing proposition ; 
And because BX isequal to XC,and AL to LC, therefore XL 

• '^- 6- is parallel'' to AB, and the triangle ABC similar to the tri- 

angle LXC : For the same reason, the triangle DEF is similar 
to RVF: And because BC is double of CX, and EF double of 
FV, therefore BC is to CX, as EF to FV: And upon BC, 
CX are described the similar and similarly situated redtilineal 
figures ABC, LXC ; and upon EF, FV, in like manner, are 
described the similar figures DEF, RVF : Therefore, as the 

* 22. 6. triangle ABC is to the triangle LXC, so'' is the triangle DEF 

to the triatigleRVF, and, by permutation, as the triangle ABC 
to the triangle DEF, so is the triangle LXC to the triangle 
R VF : And because the planes ABC, OMN, as also theplanes 

e 15. 11. DEF, STY are parallel, the perpendiculars drawn from the 
points G, H to the bases ABC, DEF, which, by the hypothe- 
sis, are equal to one another, shall be cut each into two equal 

" 17. n. "iparts by the planes OMN, STY, because the straight lines 
GC, HF are cut into two equal parts in the points N, Y by 
the same planes : Therefore theprismsLXCOMN, R VEST Y 
are of the same altitude j and therefore> as the base LXC to 

,tlic 



OF EUCLID. 



^53 



,32. 



the base RVF ; that is, as the triangle ABC to the triangle Book xii. 
DEF, so* is the prism having the triangle LXC for its base, ^ 
and OMN the triangle opposite to it, to the prism of which the ^ 
base is the triangle RVF, and the opposite triangle STY : ■ 
And because the two prisms in the pyramid ABCG are equal 
to one another, and also the two prisms in the pyramid DEFH 
equal to one another, as the prism of which the base is the 
parallelogram KBXL and opposite side MO, to the prism 
having the triangle LXC for its base, and OMN the triangle 
opposite to it ; so is the prism of which the base** is the paralle- * i. 
logram, PEV^R, and opposite side TS, to the prism of which 
the base is the triangle RVF, and opposite triangles STY. 
Therefore,componendo,as the prisms KBXLMO,LXCOMN 





together are unto the prism LXCOMN ; so are the prisms 
PEVRTS, RVFSTY to the prism R VFSTY : And permu- 
taado, as the prisms KBXLMO, LXCOMN are to the prisms 
PEVRTS, RVFSTY ; so is the prism LXCOMN to the 
prism RVFSTY : But as the prism LXCOMN to the prism 
RVFSTY, so is, as has been proved, the base ABC to the base 
DEF: Therefore, as the base ABC to the base DEF, so are 
the two prisms in the pyramid ABCG to the two prisms in 
the pyramid DEFH : And likewise if the pyramid? now made, 
for example, the two OMNG, ST YH be divided iri the same 
manner ; as the base OMN is to the base STY, so shall the 
two prisms in the pyramid OMNG be to the two prisms in 
the pyramid ST YH : But the base OiMN is to the base STY, 
as the base ABC to the base DEF ; therefore, as the base ABC 
to the base DEF, so are the two prisms in the pyramid ABCG 

to 



254 THE ELEMENTS 

Book xii. ^q (he two prisms in the pyramid DEFH ; and so are thctw© 
^ prisms in the pyramid OMNG to the two prisms in the py- 
ramid STYH ; and so are all four to all four: And the same 
thing may be shewn of the prisms made by dividing the pyra- 
mids AKLO and DPRS, and of all made by the same num- 
ber of divisions. Q. E. D. 

PROP. V. THEOR. 

JL VRAM JDS of the same altitude which have tri- 
angukr bases, are to one another as their bases. 

Let the pyramids of which the triangles ABC, DEF are the 
bases, and of which the vertices are the points G, H, be of the 
same altitude : As the base ABC to the base DEF, so is the 
pyramid ABCG to the pyramid DEFH. 

For, if it be not so, the base ABC must be to the base DEF, 
as the pyramid ABCG to a solid either less than the pyramid 
DEFH, or greater than it*. First, let it be to a solid less than 
it, viz. to the solid Q_: And divide the pyramid DEFH into 
two equal 'pyramids, similar to the whole, and into two equal 

•:.12. prisms. Therefore these two prisms are greater^ than the half 
of the whol? pyramid. And again, let the pyramids made by 
this division be in like manner divided, and so on, until the 
pyramids which -remain undivided in the pyramid DEFH be, 
all of them together, less than the excess of the pyramid DEFH 
above the solid Q_: Let these, for example, be the pyramids 
DPRS, STYH : Therefore the prisms, which make the rest of 
the pyramid DEFH, are greater than the solid Q^: Divide, 
likewise the pyramid ABCG in the same manner, and into as 
many parts, as the pyramid DEFH : Therefore as the base 
• *•• ABC to the base DEF, so'' are the prisms in the pyramid 
ABCG to the prisms in the pyramid DEFH : But as the base 
ABC tothe base DEF, so, by hypothesis, is the pyramid ABCG 
to the solid Q_; and therefore, as the pyramid ABCG to the 
solid Q_, so are the prisms in the pyramid ABCG to the prisms 
in the pyramid DKFH: But the pyramid ABCG is greater 

" ^^- ^- than the prisms contained in it ; wherefore*^ also the solid Qjs 
greater than the prisms in the pyramid DEFH. But it is also 
less, which is impossible. Therefore the bas6 ABC is not to 

the 

♦ This may be explained ll-.e same way as at the note f iu Proposition 2. in th^ 
Kke cajc. 



OF EUCLID. 



255 



the base DEF, as the pyramid ABCG to any solid which is BooxXil. 
less than the pyramid DEFH. In the same manner it may v^v*^' 
be demonstrated, that the base DEF is not to the base ABC, 
as the pyramid DEFH to any solid which Is less than the pyra- 
mid ABCG. Nor can the base ABC be to the base DEF, as 
the pyramid ABCG to any solid which is greater than the 
pyramid DEFH. For if it be possible, let it be so to a greater, 
viz. thesolid Z. And because the base ABC ist»the base DBF 
as the pyramid ABCG to thesolidZ ; by inversion, as the base 
DEF to the base ABC, so is the solid Z to the pyramid ABCG. 
But as thesolid Z is to the pyramid ABCG, so is tfte pyramid 





K 


\ 




a 






\ 


\ 








\ • 


\ 


2 




N 


\ 



DEFH to some solid*, which mu>.t be Icss^ than the pyramid * ^^• 
ABCG, because the solid Z is greater than the pyramid 
DEFH, And therefore, as the base DEF to the base ABC, 
io is the pyramid DEFH to a solid less than the pyramid 
ABCG ; the contrary to which has been proved. Therefore 
the base ABC is not to the base DEFH, as the pyramid ABCG 
to any solid which is greater than the pyramid DEFH. And 
it has been proved, that neither is the base ABC to the base 
DEF, as the pyramid ABCG to any solid which is less than 
the pyramid DEFH. Therefore, as the base ABC is to the 
kase DEF, so is the pyramid ABCG to the pyramid DEFH, 
Wherefore pyramids, &c. Q_ E, D. 



* This asay be upbioed the sane vgy as the Tike at fU iMrk f i» frf. 2. 



256 



Book XII. 



THE ELEMENTS 



SeeN. 



•5. 12. 



PROP. VI. THEOR. 



i Y RAM I Ds of the same altitude which have poly- 
gons for their bases, are to one another as their 
bases. 



Letthepyramids which havethepolygonsABCDE,FGHKL 
for their bases, and their vertices in the points M, N be of the 
same altitude : As the base ABCDE to the base FGHKL, so 
is the pyramid ABCDEiM to the pyramid FGHKLN. 

Divide the base ABCDE into the triangles ABC, ACD, 
ADE ; and the base FGHKL into the triangles fGH,FHK, 
FKL : And upon the bases ABC, ACD, ADE let there be as 
many pyramids of which the common vertex is the point M, 
and upon the remaining bases as many pyramids having their 
common vertex in the point N: Therefore since the triangle 
ABC is to the triangle FGH, as* the pyramid ABCM to the 
pyramid FGHN ; and the triangle ACD to thetriangle FGH*, 
as the pyramid ACDM to the pyramid FGHN ; and also the 



»2 Co 





D F 



triangle ADE to the triangle FGH, as the pyramid ADEMto 
the pyramid FGHN ; as all the first antecedents to their com- 
mon consequent ; so'' are all the other antecedents to their com- 
mon consequent ; that is, as the base AfiCDE to the base 
FGH, so is the pyramid ABCDEM to the pyramid FGHN ; 
And for the same reason, as the base FGHKL tothe base FGH, 
so is the pyramid FGHKLN tothe pyramid FGHN : And, by 
inversion, as the base FGH to the base FGHKL, so is the py- 
ramid FGHN tothepyramidFGHKLN; Then, because as the 
base ABCDE to the base FGH, so is the pyramid ABCDEM 
to the pyrarhid FGHN ; and as the base FGH to the base 
FGHKL,sois the pyramid FGHN to the pyramid FGHKLN; 

therefore 



OF EUCLID. 257 

therefore, ex jequali% as the base ABCDE to the base '^ookXII. 
FGHKL, so the pyramid ABCDEM to the pyramid ^"^^^ 
FGHKLN. Therefore pyramids, &c. Q^E. D. 



PROP. VII. THEOR. 

xirfVERY prism having a triangular base may be 
divided into three pyramids that have triangular 
bases, and are equal to one another. 

Let there be a prism of which the base is the triangle ABC, 
and let DEF be the triangle opposite to it : Tlie prism 
ABCDEF may be divided into three equal pyramids having 
triangular bases. 

Join BD, EC, CD ; and because ABED is a parallelogram 
of which BD is the diameter, the triangle ABD is equal* to »si. ]. 
the triangle EBD ; therefore the pyramid of which the base is 
the triangle ABD, and veitex the point C, is equal*" to the ^ 5. 12. 
pyramid of which the base is the triangle EBD, and vertex the 
point C : But this pyramid is the same with the pyramid the 
base of which is the triangle EBC, and vertex the point D ; 
for they are contained by the same planes : Therefore the 
pyramid of which the base is the triangle ABD, and vertex the 
point C, is equal to the pyramid, the base of which is the tri- 
angle EBC, and vertex the point D : Again, because FCBE is 
a parallelogram of which the diameter is 
CE, the triangle ECF is equal* to the 
triangle FCB j therefore the pyramid of 
which the base is the triangle ECB, and 
vertex the point D, is equal to the pyra- 
mid the base of which is the triangle ECF, 
and vertex the point D : But the pyramid 
of which the base is the triangle ECB, 
and vertex the point D, has been proved 
equal to the pyramid of which the base is 
the triansrle ABD, and vertex the point C. Therefore the 
prism ABCDEF is divided into three equal pyramids having tri- 
angular bases, viz. into the pyramids ABDC, EBDC, ECED: 
And because the pyramid of which the base is the triangle ABD, 
and vertex the point C, is the same with the pyramid of which 
the base is the triangle ABC, and vertex the point D, for they 
are contained by the same planesj and that the pyramid of which 
the base is the triangle ABD, and vertex the point C, has been 

S demonstrated ~ 




25^ 



THE ELEMENTS 



Book XII. demonstrated to be a third part of the prism, the base of which 
^'^"^^'^'^ is the triangle ABC, and to which DEF is the opposite 
triangle ; therefore the pyramid of which the base is the tri- 
angle ABC, and vertex the point D, is the third part of the 
prism which has the same base, viz. the triangle ABC, and 
DEF is the opposite triangle. Q. E. D. 

Cor. I. From this it is manifest, that every pyramid is the 
third part of a prism which has the same base, and is of an 
equal altitude with it; for if the base of the prism be any other 
figure than a triangle, it may be divided into prisms having 
triangular bases. 

Cor. 2. Prisms of equal altitudes are to one another as 
their bases ; because the pyramids upon the same bases, and of 
= 6. 12. the same altitude, are«= to one another as their bases. 

PROP. VIII. THEOR. 

oIMILAR pyramids, having triangular bases, are 
one to another in the triplicate ratio of that of their 
homologous sides. 

Let the pyramids having the triangles ABC, DEF for their 
bases, and the points G, H for their vertices, be similar, and 
similarly situated ; the pyramid ABCG has to the pyramid 
DEFH, the triplicate ratio of that which the side BC has to 
the homologous side EF. 

Complete the parallelograms ABCM, GBCN, ABGK, and 
the solid parallelopiped BGML contained by these planes and 




X 




those opposite to them: And, in like manner, complete the solid 
parallelopiped EHPO contained by the three parallelograms 
DEFP, HEFR, DEHX, and those opposite to them : And 

because 



OF EUCLID. 259 

because the pyramid ABCG is similar to the pyramid DEFH, ^°»^ xii. 
the angle ABC is equal * to the angle DEF, and the angle .YTd^Tii 
GBC to the angle HEF and ABG to DEH : And AB is'' to";. d«f.<. 
BC, as DE to EF; that is, the sides about the equal angles 
are proportionals ; wherefore the parallelogram BM is similar 
toEP : For the same reason, the parallelogram BN is similar 
to ER, and BK to EX : Therefore the three parallelograms 
BM,BN, BKl are similar to the three EP, ER, EX : But the ' 
three BM, BN, BK, are equal and similar*^ to the three which c 24. 11. 
are opposite to them, and the three EP, ER, EX equal and 
similar to the three opposite to them : Wherefore the solids 
BGML, EHPO are contained by the same number of similar 
planes ; and their solid angles are equal"* ; and therefore thed b ^ 
solid BGML is similarHothe solid EHPO : But similar solid 
parallelopipeds have the triplicate •" ratio of that which theiresg, ]]_ 
homologous sides have : Therefore the solid BGML has to 
the solid EHPO the triplicate ratio of that which the side BC 
has to the homologous side EF : But as the solid BGML is to 
the solid EHPO, so is '^ the pyramid ABCG to the pyramid f 15. 5. 
DEFH ; because the pyramids are the sixth part of the solids, 
since the prism, which is the half§ of the solid parallelopiped, e2S. ii. 
is triple ^ of the pyramid. Wherefore likewise the pyramid » 7. 12. 
ABCG has to the pyramid DEFH, the triplicate ratio of that 
which BC has to the homologous side EF, Q. E. D. 

Cor. From this it is evident, that similar pyramids which Sec J«'. 
have multangular bases, are likewise to one another in the 
triplicate ratio of their homologous sides : For they may be 
divided into similar pyramids having triangular bases, because 
the similar polygons, which are their bases, may be divided 
into the same number of similar triangles homologous to the 
whole polygons ; therefore as one of the triangular pyramids 
in the first multangular pyramid is to one of the triangular 
pyramids in the other, so are all the triangular pyramids in the 
first to all the triangular pyramids in the other j that is, so is 
the first multangular pyramid to the other : But one triangular 
pyramid is to its similar triangular pyramid, in the triplicate 
ratio of their homologous sides ; and therefore the first mul- 
tangular pyramid has to the other, the triplicate ratio of that 
which one of the sides of the first has to the homologous 
•ides of the other. 



S2 



:6o 



THE elem:e:nts 



;.X XII- 



PROP. IX. THEOR. 

1 HE bases aiicl altitudes of equal pyramids.liav- 
ing triangular bases are reciprocally proportional : 
And triangular pyrajiiids of which the bases and 
altitudes are reciprocally proportional, are equal to 
one another. 

Let the pyramids of which the triangles ABC, DEF, are 
the bases, and which have their vertices in the points G, H, be 
equal to one another ; The bases and altitudes of the pyramids 
ABCG, DEFH are reciprocally proportional, viz. the base 
ABC is to the base DEF, as the altitude of the pyramid 
.DEFFI to the altitude of the pyramid ABCG. 

Complete the parallelograms AC, AG, GC, DF, DH, HF ; 
and- the solid^ parallelopipeds BGML, FHPO, contained by 





•tliese planes and those opposite to them : And because the 
pyramid ABCG is eq«al to the pyramid DEFH, and that the 
solid BGML is sextuple of the pyramid ABCG, and the solid 
EHPO sextuple of the pyramid DEFH ; therefore the solid 

' 1. Ax.5. BGML is equal * to the solid EHPO : Put the bases and alti- 
tudes of equal solid parallelopipeds are reciprocally propor- 

*3\. \\. tionalb; therefore as the base BM to the base FP, so is the- al- 
titude of the solid EHPO to the altitude of the solid BGML : 
But as the base BM to the base EP, so is"^ the triangle ABC 
to the triangle DEF ; therefore as the triangle ABC to the tri- 
angle DEF, so is the altitude of the solid EHPO to the alti- 
tude of the solid BGML: But the altitude of the solid EHPO 
is the same with the altitude of the pyramid DEFH ; and the 
altitude of the s»lid BGML is the same with the altitude of the 

pyramid 



' 15. 5. 



OF EUCLID. 261 

pyramid ABCG: Therefore, as the base ABC to the base DEF, 8ooK^?tII. 
so is' the altitude of the pyramid DEFH to the altitude of the ''*''^''**^ 
pyramid ABCG : Wherefore the bases and altitudes of the 
pyramids ABCG, DEFH are reciprocally proportional. 

Again, let the bases and altitudes of the pyramids ABCG, 
DEFH be reciprocally proportional, viz. the base ABC to the 
base DEF, as the altitude of the pyramid DEFH to the alti- 
tude of the pyramid ABCG : The pyramid ABCG is equal 
to the pyramid DEFH. 

The same construction bejng made, because as the base ABC 
to th J base DEF, so is the altitude of the pyramid DEFH to 
the altitude of the pyramid ABCG : And as the base ABC to 
the base DEF, so is the parallelogram BM to the parallelo- 
gram EP ; therefore the parallelogram BM is to EP, as the 
altitude of the pyramid DEFH to the altitude of the pyramid 
ABCG : But the altitude of the pyramid DEFH is the same 
with the altitude of the solid parallelopiped EHPOi and the 
altitude of the pyramid ABCG is the same with the altitude 
of the solid parallelopiped BGML : As, therefore, the base 
BM to the base EP, so is the altitude of the solid parallelopiped 
EHPO to the altitude of the solid parallelopiped BGML. 
But solid parallelopipeds having; their bases and altitudes reci- 
procally proportional, are equal*" to one another. Therefore " 34. 11. 
the solid parallelopiped BGML is equal to the solid parallelo- 
piped EHPO. And the pyramid ABCG is the sixth part of 
the solid BGML, and the pyramid DEFfI is the sixth part of 
the solid EHPO. Therefore the pyramid ABCG is equal to 
the pyramid DEFH. Therefore the bases, &c. Q. E. D. 



PROP. X. THEOR. 

JiiVERY cone is the third part of acyhnder which 
has the same hase, and is of an equal altitude with 
it. 

Let a cone have the same base with a cylinder, viz. the cir- 
cle ABCD, and the same altitude. The cone is the third part 
of the cylinder ; that is, the cylinder is triple of the cone. 
If the cylinder be not triple of the cone, it must either be 
■' greater than the triple, or less than it. First, Let it be greater 
"^f than the triple; and describe the square ABCD in the circle: 
this square is greater than the half of the circl« ABCD*. 

S 3 Upoiv 

* As \rai «hewii in Pro^. 2 gf tlu» B«ok. 



262 THE ELEMENTS 

Book XII. Upon the Square ABCDere£l a prism of the same altitiade with 
^^^"^ the cylinder ; this prism is greater than half of the cylinder j 
because if a square be described about the circle, and a prism 
ereiSled upon the square, of the same altitude with the cylinder, 
the inscribed square is half of that circumscribed ; and upon 
these square bases are ere6led solid parallelopipeds, viz. the 
prisms of the same altitude ; therefore ' the prism upoh the 
square ABCD is the half of the prism upon the square described 
about the circle : Because they are to one another as their 
> 3* 1 1, bases* : And the cylinder is less than the prism upon the square 
described about the circle ABCD: Therefore the prism upon 
the square ABCD of the same altitude with the cylinder, is 
greater than half of the cylinder. Bisedt the circumferences 
AB, BC, CD, DA in the points E, F, G, H ; and join AE, 
EB, BF, FC, CG, GD, DH, HA : Then, each of the triangles 
AEB, BFC, CGD, DHA is greater than the half of the seg- 
ment of the circle in which it stands, 
as was shewn in Prop. 2. of this 
Book. Ere£t prisms upon each of 
these triangles of the same altitude 
with the cylinder; each- of these 
prisms is greater than half of the seg- 
ment of the cylinder in which it is ; 
because if, through the points E, F, 
G, H, parallels be drawn to AB, BC, 
CD, DA, and parallelograms be 
completed upon the same AB, BC, 
CD, DA, and solid parallelopipeds 
be eredled upon the parallelograms ; the prisms upon the tri- 
angles AEB, BFC, CGD, DHA are the halves of the solid 
► '2 Cor. parallelopipeds''. And the segments o( the cylinder which arc 
7. u. upon the segments of the circle cut ofFby AB, BC, CD, DA, 
are less than the solid parallelopipeds which contai.n them. 
Therefore the prism3 upon the triangles AEB, BFC, CGD, 
DHA, are greater than half of the segments of the cylinder in 
which they are j therefore, if each of the circumferences be 
divided into two equal parts, and straight lines be drawn from 
the points of division to the extremities of the circumferences, 
and upon the triangles thus made, prisms be eredled of the same 
altitude with the cylinder, and so on, there must at length 
Lcmms. remain some segments of the cylinder which together are less,'^ 
than the excess of the cylinder" above the triple of the cone, 
Let them be those upon the segments of the circle AE, EB, BF. 

FC, 




OF EUCLID. 



263 



19. 



FC, CG, GD, DH, HA. Therefore the rest of the cylin- BoorXII. 
der, that is, the prism of which the base is the polygon 
AEBFCGDH, and of which the altitude is the same with that 
of the cylinder, is greater than the triple of the cone : But this 
prism is triple"* of the pyramid upon the same base, of which * 1 Cor. T. 
the vertex is the same with the vertex of the cone; therefore 
the pyramid upon the base AEBFCGDH, having the same 
vertex with the cone, is greater than the cone, of which the 
base is the circle ABCD : But it is also less, for the pyramid is 
contained within the cone ; which is impossible. Nor can the 
cylinder be less than the triple of the cone. Let it be less, if 
possible ; therefore, inversely, the cone is <Treater than the third 
part of the cylinder. In the circle ABCD describe a square ; 
this square is greater than the half of the circle; And upon the 
square ABCD ereft a p) ramid, having the same vertex with the 
cone: this pyramid is greater than the halfof the cone j because, 
as was before demonstrated, if a square be described about the 
circle, the square ABCD is the 
half of it; and if upon these squares 
there be erected solid parallelo- 
pipeds of thesame altitude wih the 
cone, which are also prisms, the 
prism upon the square ABCD 
shall be the half of that which is 
upon the square, described about 
the circle; for they are to one 
another as their bases'^ ; as are also 
the third parts of them : Therefore 
the pyramid, the base of which is 

the square ABCD, is half of the pyramid upon the square de- 
scribed about the circle : But this last pyramid is greater than 
the cone which it contains ; therefore the pyramid upon the 
square ABCD, having the same vertex with the cone, is greater 
than the half of the cone. Bisect the circumferences AB, 
BC ; CD, DA in the points E, F, G, H, and join AE, EB, 
BF, FC, CG, GD, DH, HA: Therefore each of the triangles 
AEB, BFC, CGD, DHA is greater than half of the segment 
of the circle in which it is ; Upon each of these triangles ereft 
pyramids having the same vertex with the cone. Therefore 
each of those pyramids is greater than the half of the segment 
of the cone in which it is, as before was demonstrated of the 
prisms and segments of the cylinder ; and thus dividing each 
©f the circumferences into two equal parts, and joining the 

S 4 points 




« 3'i. 11. 



264 



THE ELEMENTS 



liooK XII. points of division and their extremities by straight lines, and 
"'^"''''^'^ upon the triangles erecting pyramids having their vertices the 
same with that of the cone, and so on, there must at length 
remain some segments of the cone, which together shall be less 
than the excess of the cone, above the third part of the cylinder. 
Let these be the segments upon AE, EB, BF, FC, CG, GD, 
DH, HA. Therefore the rest of 
the cone, that is, the pyramid, of 
which the base is the polygon 
AEBFCGDH, and of which the 
vertex is the same with that of the 
cone, is greater than the third part 
of the cylinder. But this pyramid 
is the third part of the prism upon 
the same base AEBFCGDH, and 
of the same altitude with the cylin- 
der. Therefore this prism is great- 
er than the cylinder of which the 

base is the circle A BCD. But it is also less, for it is contained 
within the cylinder ; which is impossible. Therefore the c'y- 
, linder is not less than the triple of the cone. And it has been 
demonstrated that neither is it greater than the triple. There- 
fore the cylinder is triple of the cone, or, the cone is the third 
part of the cylinder. Wherefore every cone, &c. C^ E. D. 




Sf« N. 



C 



PROP. XL THEOR. 



ONES and cylinders of the same altitude, are to 
one another as their bases. 

Let the cones and cylinders, of which the bases are the cir- 
cles ABCD,EFGH, and the axes KL, MN, and AC, EG the 
diameters of their bases, be of the same altitude. As the circle 
A BC D to the circle EFGH, so is the cone AL to the cone EN. 

If it be not so, let the circle ABCD be to the circle EFGH, 
as the cone AL to some solid either less than the cone EN, or 
greater than it. First, let it be to a solid less than EN, viz. to 
the solid X ; and let Z be the solid which is equal to the excess 
of the cone EN above the solid X ; therefore the cone EN 
is equal to the solids X, Z together. In the circle EFGH 
describe the square EFGH, therefore this square is greater than 
the half of the circle : Upon the square EFGH eredl a pyra- 
mid of the same altitude with the cone ; this pyramid is 
greaiter than half of the cone. For, if a square be de- 
scribed about the circle, and a pyramid be ere^ed upon it, 

having 



OF EUCLID. 



265 



having the same vertex with the cone*, the pyramid inscribed book xii. 
in the cone is half to the pyramid circumscribed about it, *^—n^^^ 
because they are to one another as their bases* : But the cone »6. 12. 
is less than the circumscribed pyramid ; therefore the pyramid 
of which the base is the square EFGH, and its vertex the same 
with that of the cone, is greater than half of the cone: Divide 
the circumferences EF, FG, GH, HE, each into two equal 
parts in the points O, P, R, S, and join EO, OF, FP, PG, 
GR, RH, HS, SE : Therefore each of the triangles EOF, 
FPG, GRH, HSE is greater than half of the segment of the 





\ 




N 




\ 




\ 




I 








X 






^ 




\ 


n 




\. 


V 




N 


\i 


z 


\ 



circle in which it is : Upon each of these triangles ere£l; a py- 
ramid having the same vertex with the cone ; each of these 
pyramids is greater than the half of the segment of the cone in 
which it is : And thus dividing each of these circumferences 
into two equal parts, and from the points of division drawing 
straight lines to the extremities of the circumference, and up- 
on each of the triangles thus made erecting pyramids, having 
the same vertex with the cone, and so on, there must at length 
remain some segments of the cone which are together less** * Lea. 
than the solid Z: Let these be the segments upon EO, OF, FP, 

PG, 

• Vertex is put in place of aUitude, which is in the Qreek, because the pjTjmid, ia 
■what follows, it supposed to be circumscribed abcjt the cone, and so must have thK 
»»inc fcrtex. Aad the ssae cbang* it waas in scne place: ibUo'Vf ing. 



1 



266 



THE ELEMENTS 



Book XII. pQ, QR, RH, HS,SE: Therefore the remainder of the cone, 
^"^ viz. thepyramidofwhichthebaseisthepolygon EOFPGRHS, 
and its vertex the same with that of the cone, is greater than 
the solid X : In the circle ABCD describe the polygon 
ATBYC VDQ^similartothe polygon EOFPGRHS,and upon 
it erect a pyramid having the same vertex w^ith the cone AL : 
And because as the square of AC is to the square of EG, so^ is 
the polygon ATBYCVDQ^to the polygon EOFPGRHS; 
and as the square of AC to the square of EG, so is*" the circle 
ABCD to the circle EFGH ; therefore the circle ABCD*^ is to 
the circle EFGH,asseepolygonATBYCVDQ to the polygon 



»1. 12. 



" 2. 12. 
<= 11.5. 




C E 




I 



f\ 




[\ 




X 




\ 




\ 






N 


\ 






\ 


\ 


z 



EOFPGRHS : But as the circle ABCD to the circle EFGH, 
so is the cone AL to the solid X: and as the polygon 
*^' ATB YC VDQ^to the polygon EOFPGRHS, sd is" the pyra- 
mid of vi^hich the base is the first of these polygons, and vertex 
L, to the pyramid of which the base is the other polygon, and 
its vertex N : Therefore, as the cone AL to the solid X, so is 
the pyramid of which the base is the polygon ATBYC VDQ^, 
and vertex L, to the pyramid the base of which is the polygon 
j4 3 EOFPGRHS, ajid vertex N : But the cone AL is greater than 
the pyramid contained in it ; therefore the solid X is greater'^ 
than the pyramid in the cone EN. But it isless, as was shewn, 

whicft 



OF EUCLID. 267 

which is absurd : Therefore the circle ABCD is not to the Book xir. 
circle EFGH, as the cone AL to any solid which is less than ^■^^•^*-' 
the cone EN. In the same manner it may be demonstrated, 
that the circle EFGH is not to the circle ABCD, as the 
cone EN to any solid less than the cone AL. Nor can the 
circle ABCD be to the circle EFGH, as the cone AL to any 
solid greater than the cone EN : For, if it be possible, let it 
be so to the solid I, which is greater than the cone EN : 
Therefore, by inversion, as the circle EFGH to the circle 
ABCD, so is the solid I to the cone AL : But as the solid I 
to the cone AL, so is the cone EN to some solid, which must 
be less» than the cone AL, because the solid I is greater than » u. 3t 
the cone EN : Therefore, as the circle EFGH is to the circle 
ABCD, so is the cone EN to a solid less than the cone AL, 
which was shewn to be impossible : Therefore the circle 
ABCD is not to the circle EFGH, as the cone AL is to any 
solid greater than the cone EN : And it has been demonstra- 
ted, that neither is the circle A^CD to the circle EFGH, as 
the cone AL to any solid less than the cone EN : Therefore 
the circle ABCD is to the circle EFGH, as the cone AL to 
the cone EN : But as the cone is to the cone, so'' is the cy- jj 5 
linder to the cylinder, because the cylinders are triple'^ of the 
cone, each to each. Therefore as the circle ABCD to the 
circle EFGH, so are the cylinders upon them of the same 
altitude. Wherefore cones and cylinders of the same altitude 
are to one another a,s their bases. Q. E. D. 



PROP. XII. THEOR. 

Similar cones and cylinders have to one ano- seeN. 
ther the triplicate ratio of that which the diameters 
of their bases have. 

Let the cones and cylinders of which the bases are the cir- 
cles ABCD, EFGH, and the diameters of the bases AC, EG, 
and KL, MN, the axis of the cones or cylinders, be similar : 
The cone of which the base is the circle ABCD, and vertex the 
point L, has to the one of which the base is the circle EFGH, 
and vertex N, the triplicate ratio of that which AC has to EG. 

For if the cone ABCDL has not to the cone EFGHN the 
triplicate ratio of that which AC has to EG, the cone ABCDL 
shall have the triplicate of that ratio to some solid Which is less 
or greater than the cone EFGHN. First, let it have it to a less, 

viz. 



= 10. 13. 



268 



THE. ELEMENTS 



Book xn. viz to the solid X. Make the same construction as in the pre- 
'"^^ ceding proposition, and it may be d-monstrated the very same 
way as in that proposition, that the pyramid of which the base 
is the polygon EOFPGRHS, and vertex N, is greater than 
the solid X. Describe also in the circle ABCD the polygon 
ATBYCVDQ similar to the polygon EOFPGRHS, upon 
which ere6l a pyramid having the same vertex with the cone ; 
and let LAQ^ be one of the triangles containing the pyra- 
mid upon the polygon ATBYCVDQ^, the vertex of which 
is L ; and let NEb be one of the triangles containing the 




pyramid upon the polygon EOFPGRHS of which the ver- 
tex is N J and join KQ^, MS : Because then the- cone 
='24def.ii. ABCDL is similar to the cone EFGHN, AC is' to EG as 
■i>i5. 5. the axis KL to the axis MN ; and as AC to EG, so* is AK 
to EM ; therefore as AK to EM, so is KL to MN ; 
and, alternately, AK to KL, as EM to MN : And the 
right angles AKL, EMN are equal j therefore the sides 
about these equal angles beirig proportionals, the triangle 
AKL is similar*^ to the triangle EMN. Again, because 
AK is to KQ,, as EM to MS, and that these sides are 

about 



« (kG. 



OF EUCLID. 269 

about equal . angles AKQj EMS, because tbese angles are, BookXII, 
each of them, the same part of four right angles at the centres 
K, IVl ; therefore the triangle AKQ_is similar^ to the triangle a g. g. 
EiVlS : AnJ because it has been shown that as AK to KI^ so 
is EM to MN, and that AK is equal to KQ_; and EM to MS, 
as QK to KL, so is SM to MN : and therefore the sides about 
the right angles QKL, SMN being proportionals, the triangle 
LKQ^is similar to the triangle NMS : and because of the 
similarity of the triangles AKL, EMN, as LA is to AK, so is 
NE to EM; and by the similarity of the triangles AKQ_, 
EMS, as KA to AQ^, so ME to ES ; ex lequali^ LA is " --• ^ 
to AQ, as NE to ES. Again, because of the similarity of the 
triangles LQK, NSM, as LQ, to QK, so NS to SM ; and 
from the similarity of the triangles KAQ_, MES, as KQ^to 
QA, so MS to SE ; ex aequali'', LQ is to Q_\, as NS to SE : 
And it was proved that QA is to AL, as SE to EN ; there- 
fore, again, ex sequali as QL to LA, so is SN to NE : Where- 
fore the triangles LQ_A, NSE, having the sides about all their 
angles proportionals, are equiangular' and similar to one'^S-fi- 
another : And therefore the pyramid of which the base is the 
triangle AKQ_, and vertex L, is similar to the pyramid the base 
of which is the triangle EMS, and vertex N, because their 
solid angles are equal** to one another, and they are contained ''B. 11. 
by the same number of similar planes: But similar pyramids 
which have- triangular bases have to one another the triplicate 
'ratio of that which their homologous sides have; therefore *s. 12. 
the pyramid AKQL has to the pyramid EMSN the triplicate 
ratio of that which AK has to EM. In the same manner, if 
straight lines be drawn from the points D, V, C,"Y, B, T, 
to K, and from the points H, R, G, P, F, O, to M, and py- 
ramids be erected upon the triangles having the same vertices 
with the cones, it may be demonstrated that each pyramid in 
the first cone has to each in the other, taking them in the same 
order, the triplicate ratio of that which the side AK hastothc 
side EM ; that is, which AC has to EG : But as one antece- 
dent to its consequent, so are all the antecedents to all the 
consequents*^ ; therefore as the pyramid AKQL to the pyra- ^ 12. 5. 
mid EMSN, so is the whole pyramid the base of which is the 
polygon DQATBYCV, and vertex L, to the whole pyramid 
of which the base is the polygon HSEOFPGR,and vertex N. 
Wherefore also the first of these two last-named pyramids has 
to the other the triplicate ratio of that which AC has to EG, 
But, by the hypothesis, the cone of which the base is the cir- 
cle ABCD, and vertex L, has to the solid X, the triplicate 
ratio of that which AC has to EG ; therefore, as ihe cone of 

which 



270 



THE ELEMENTS 



Book xri. which the base is the circle ABCD, and vertex L, is to the 
^"""'''^'^ solid X, so is the pyramid the base of which is the polygon 
DQA i'-BYCV, and vertex L, to the pyramid the base of which 
is the polygon HSEOFPGR and vertex N : But the said cone 
^ is greater than the pyramid contained in it, therefore the solid 

X is greater^ than the pyramid, the base of which is the poly- 
gon HSEOFPGR, and vertex N ; but it is also less, which is 
impossible : therefore the cone, of which the base is the circle 



u 




\ 


\ 


X 


\ 




ABCD and vertex L, has not to any solid which is less than the 
cone of which the base is thecircle EFGH and vertexN,the tri- 
plicate ratio of that which AC has to EG. In the same manner 
it may be demonstrated, that neither has the cone EFGHN to 
any solid which is less than the cone ABCDL, the triplicate 
ratio of that which EG has to AC. Nor can the cone A BCDL 
have to any solid which is greater than the cone EFGHN, the j 
triplicate ratio of that which AC has to EG : For, if it be pos- 
sible, let it have it to a greater, viz. to the solid Z : Therefore, 
inversely, the solid Z has to the cone ABCDL, the triplicate 
ratio of that which EG has to AC : But as the solid Z is to 

the 



OF EUCLID. 



271 



thecone ABCDL, soistheconeEFGHN to some solid," which BookXil 
tnust be less* than the cone ABCDL, because the solid Z is , ^^ ^ 
greater than the cone EFGHN : Therefore the cone EFGHN 
has to a solid which is less than the cone ABCDL, the tripli- 
cate ratio of that which EG has to AC, which was demon- 
strated to be impossible : therefore the cone ABCDL has not 
to any solid greater than the cone EFGHN, the triplicate ra- 
tio of that which AC has to EG ; and it was demonstrated, that 
it could not have that ratio to any solid less thaa the cone 
EFGHN : Therefore the cone ABCDL has to the cone 
EFGHN, the triplicate ratio of that which AC has to EG : 
But as the cone is to the cone, so'' the cylinder to the cylinder; ' 
for every cone is the third part of the cylinder upon the same 
base, and of the same altitude ; Therefore also the cylinder 
has to the cylinder, the triplicate ratio of that v/hich AC has 
to EG: Wherefore similar cones, Sec. Q. E. D. 



15. 5. 



PROP. Xin. THEOR. 

-IF a cylinder be cut by a plane parallel to its oppo- See n. 
site planes, or bases ; it divides the cylinder into two 
cylinders, one of which is to the other as the axis 
of the first to the axis of the other. 

Let the cylinder AD be cut by the 
plane GH parallel to the opposite 
planes AB, CD, meeting the axis 
EF in the point K, and let the line 
GH be the common section of the 
plane GHand the surface of the cylin- 
der AD : Let AEFC be the paralle- 
logram in any position of it, by the 
revolution of which about the straight 
line EF the cylinder AD is described : 
and let GK be the common section 
of the plane GH, and the plane 
AEFC: And because the parallel 
planes AB, GH, are cut by the plane 
AEKG, AE, KG, their commoa 
sections with it are parallel* ; where- 
fore AK is a parallelogram, and GK 
equal to FA the straight line from 
the centre of the circle AB : For the 
same reason, each of the straight lines 



K 



A 




S 
B 

H 
D 
Y 

drawn 



16, II, 



272 
EooK'XlI, 



•15. def.l 



* 11.12. 



THE ELEMENTS 

drawn from the point K to the line GH may be proved to be 
equal to those which are drawn from the centre of the circle AB 
to its circumference, and are therefore all equal to one another. 
Therefore the line GH is the circumference of a circle* j of 
which the centre is the point K : Therefore the plane GH 
divides the cylinder AD into the cylinders AH, GD ; for they 
are the same which would be described by the revolution of 
the parallelograms AK, GF, about the straight lines EK, KF: 
And it is to be shewn, that the cylinder AH is to the cylin- 
der HC, as the axis EK to the axis KF. 

Produce the axis EF both ways ; and take any number of 
Straight lines EN, NL, each equal to EK } and any number 
EX, XM, each equal to tK ; and let 
planes parallel toAB, CD pass through 
the points L, N, X, M : Therefore 
the common sections of these planes 
with the cylinder produced are circles 
the centres of which are the points 
L, N, X, M, as was proved of the 
plane GH ; and these planes cut off 
the cylinders PR, RB, DT, TQ : 
And because the axes LN, NE, EK 
are all equal; therefore the cylinders 
PR, RB, BG are'' to one another as 
their bases ; but their bases are equal, 
and therefore the cylinders PR, RB, 
BG are equal : And because the axes 
LN, NE, EK are equal to one ano- 
ther, as also the cylinders PR, RB, BG,, 
and that there are as many axes as 
cylinders ; therefore, whatever multiple 
the axis KL is of the axis KE, the same 
multiple is the cylinder PG of the cy- 
linder GB: ¥oT the same reason, whatever multiple the axis 
MK is of the axis KF, the same multiple is the cylinder QG 
of the cylinder GD : And if the axis KL be equal to the axis 
KM, the cylinder PG is equal to the cylinder GQ ; and if the 
axis KL be greater than the axis KM, the cylinder PG is 
greater than the cylinder QG ; and if less, less : Since, there- 
fore there are four magnitudes, viz. the axis EK, KF, and 
the cylinders BG, GD, and that of the axis EK and cylinder 
JJG, there has been taken any equimultiples whatever,viz. the 

axis 




OF EUCLID. 



273 



axis KL and cylinder PG ; and of the axis KF and cylinder Book xil. 
GD, any equimultiples whatever, viz. the axis KM an<? ^^^^^"""^ ■ 
cylinder GQ; arid it has been demonstrated, if the axis KLbe 
greater than the axis KM, the cylinder PG is greater than the 
cylinder GQ ; and if equal, equal ; and if less, less: Therefore 
•"the axis EK is to the axis KF, as the cylinder BG to the«5. def. r 
cylinder GD. Wherefore, if a cylinder, &c. Q. E. D. 



PROP. XIV. THEOR. 



v^ONES and cylinders upon equal bases are to 
one another as their altitudes. 

Let the cylinders EB, FD be upon the equal bases AB, 
CD : As the cylinder EB to the cylinder FD, so is the axis 
GH to the axis KL, 

Produce the axis KL to the point N, and make LN equal 
to the axis GH, and let CM be a cylinder of which the base is 
CD, and axis LN j and because the cylinders EB, CM have 
the same altitude, they are to one another as their bases' : But * n. ]o^ 
their bases are equal, therefore also the cylinders EB, CM are 
equah And because the cylin- 



der FM is cut by the plane 
CD parallel to its opposite 
planes, as the cylinder CM to 
the cylinder FD so is* the 
axis LN to the axis KL. But 
the cylinder CM is equal to 
the cylinder EB, and the axis 
LN to the axis GH : There- 
fore as the cylinder 'EB to the 
cylinder FD, so is the axis 
GH to the axis KL : And as 
the cylinder EB to the cylinder FD, so is«= the cone ABG to c 15. 5^ 
the cone CDK, because the cylinders are triple' of the cones : 4 10. ig. 
Therefore also the axis GH is to the axis KL, as the cone 
ABG to the cone CDK, and the cylinder EB to the cylinder 
FD. Wherefore cones, &c, Q^ E. D. 




k 13. 12.- 



274 

Book XIl. 



THE ELEMENTS 



ie«N. 



PROP. XV. THEOR. 

1 HE bases and altitudes of equal cones and cylin- 
ders, are reciprocally proportional ; and if the bases 
and altitudes be reciprocally proportional, the cones 
and cylinders are equal to one another. 

Let the circles A BCD, EFGH, the diameters of which are 
AC, EG, be the bases, and KL, MN the axis, as also the al- 
titudes, of equal cones and cylinders ; and let ALC, ENG be 
the cones, and AX, EO the cylinders : The bases and alti- 
tudes of the cylinders AX, EO are reciprocally proportional ; 
that is, as the base ABCD to the base EFGfi, so is the alti- 
tude MN to the altitude KL. 

Either the altitude MN is equal to the altitude KL, or these 
altitudes are not equal. First, let them be equal ; and the cylin- 
ders AX, EO being also equal, and cones and cylinders of the 
-»n. 12. same altitude being to one another as their bases*, therefore 

b A. 5. the base ABCD is equal'' to the base EFGH ; and as the 
base ABCD is to the base EFGH, so is the altitude MN to 
the altitude KL. 
But let the alti- 
tudes KL, MN, 
be unequal, and 
MN the greater 
of the two, and 
from M'N take 
MP equal to KL, 
and through the 
point P cut the 
cylinder EO by 
the plane TYS, 
parallel to the 
opposite planes of the circles EFGH, RO ; therefore the 
common sedtion of the plane TYS and the cylinder EO is 
a circle, and consequently ES is a cylinder, the base of which 
is the circle EFGH, and altitude MP : And because the cy- 
linder AX is equal to the cylinder EO, as AX is to the cylin- 

' '• ^' der ES, so « is the cylinder EO to the same ES : But as the 
cylinder AX to the cylinder ES, so* is the base ABCD to 
the base EFGH ; for the cylinders AX, ES are of the same 

* 13. 12. altitude ; and as the cylinder EO to the cylinder ES, so* is 
the altitude MN to the altitude MP, because the cylinder 

EO 





OF EUCLID. 275 

EO is cut by the plane TYS parallel to its opposite planes. 2=°^ ^^^• 
Therefore as the base ABCD to the base EFGH, so is the '*'*^***^ 
altitude MN to the altitude MP: But MP is equal to the al- 
titude KL ; wherefore as the base ABCD to the base EFGH, 
so is the altitude iMN to the altitude KL ; that is, the bases 
and altitudes of the equal cylinders AX, EO are reciprocally 
proportional. 

But let the bases and altitudes of the cylinders AX, EO, be 
reciprocally proportional, viz. the base ABCD to the base 
EFGH, as the altitude MN to the altitude KL : The cylin- 
der AX is equal to the cylinder EO. 

First, Let the "base ABCD be equal to the base EFGH ; 
then because as the base ABCD is to the base EFGH, so is 
the altitude MN to the altitude KL ; MN is equal** to KL, "A. 5. 
and therefore the cylinder AX is equal' to the cylinder EO. » n..i2. 

But let the bases ABCD, EFGH be unequal, and let 
ABCD be the greater ; and because, as ABCD is to the base 
EFGH, so is the altitude MN to the altitude KL j therefore 
MN is greater* than KL. Then, the same construction 
being made as before, because as the base ARCD to the base 
EFGH, so is the altitude MN to the altitude KL ; and because 
the altitude KL is equal to the altitude MP ; therefore the 
base ABCD is* to the base EFGH, as the cylinder AX to the 
cylinder ES ; and as the altitude MN to the altitude MP or 
KL, so is the cylinder EO to the cylinder ES : Therefore the 
cylinder AX is to the cylinder ES, as the cylinder EO is to • 
the same ES : Whence the cylinder AX is equal to the cylin- 
der EO i and the same reasoning holds in cones. Q^ E. D. ^ 



PROP. XVL PROB. 

1 O describe in the greater of two circles' that 
have the same centre, a polygon of an even number 
of equal sides, that shall not meet the lesser circle. 

Let ABCD, EFGH be two given cfrcles having the same 
centre K : It is required to inscribe in the greater circle 
ABCD, a polygon of an even number of equal sides, that 
shall not meet the lesser circle. 

Through the centre K draw the straight iine BD, and from 
%he point G, where it meets the circuiiiferencei of tfee lesser 

T 2 circle. 



276 



THE ELEMENTS 



"16. 3. 



Lemn 



Book XFt. circle, draw GA at right angles to BD, and produce it to C ; 
therefore AC touches'* the circle EFGH : Then, if the circum- 
ference BAD be bise6led, and the half of it be again bisedled, 
and so on, there must at length remain a circumference less^ 
than AD : Let this be LD ; and 
from the point L draw LM per- 
pendicular to BD, and produce 
it to N i and join LD, DN. 
Therefore LD is equal to DN j 
and because LN is parallel to AC, 
and that AC touches the circle 
EFGH i therefore LN does not 
meet the circle EFGH. And 
much less shall the straight lines 
LD, DN meet the circle E FGH ; 

So that if straight lines equal to LD be applied in thecircle 
ABCD from the point L around to N, there shall be described 
in the circle a polygon of an even number of equal sides not 
meeting the lesser circle. Which was to be done. 




«2S.3, 



LEMMA IL 

If two trapeziums ABCD, EFGH be inscribed 
in the circles, the centres of which are the points 
K, L ; and if the sides AB, DC be parallel, as also 
EF, HG ; and the other four sides AD, BC, EH, 
FG, be all equal to one another ; but the side AB 
greater than EF, and DC greater than HG. The 
straight line KA from the centre of the circle in 
which the greater sides are, is greater than the 
straight line LE drawn from the centre to the cir- 
cumference of the other circle. 

If it be possible, let KA be not greater than LE ; then K A 
must be either equal to it, or less. First, let KA be equal to 
LE : Therefore, because in two equal circles AD,BC, in the 
©ne, are equal to EH, FG in the other, the circumferences 
AD, BC are equaU to the circumferences EH, FG; but 
because the straight lines AB,DC are respedlively greater than 
EF, GH, the circumferences AB, DC are greater than EF, 
HG : Therefore the whole circumference ABCD is greater 
(^an the whole EFGH -, but it is also equal to it, which is 

inpossible : 



OF EUCLID. 



277 



impossible: Therefore the straight line KA is not equal to BookXH. 
LE. ^ '-"^'^ 

But let KA be less than LE, and make LM equal toKA, 
and from the centre L, and distance LM, describe the circle 
MNOP, meeting the straight lines LE, LP, LG, LH, in M, 
N, O, P J and join MN, NO, OP, PM, which are respectively 
parallel '^ to and less than EF, FG, GH, HE : Then > 2. 5. 
because EH is greater than MP, AD is greater than MP j and 




the circles ABCD, MNOP are equal ; therefore the circum- 
ference AD is greater than MP ; for the same reason, tha cir- 
cumference DC is greater than NO ; and because the straight 
line AB is greater than EF, which is greater than MN, much 
more is AB greater than MN : Therefore the circumference 
AB is greater than MN j and, for the same reason, the cir- 
cumference DC is greater than PO.: Therefore the whole 
circumference ABCD is greater than the whole MNOP j but 
it is likewise equal to it, which is impossible : Therefor-- KA 
is not less than LE : nor is it equal to it j the straight line 
KA must therefore be greater than LE. Q^ E. D. 

Cor. And if there be an isosceles triangle, the sides of which 
are equal to AD, BC, but its base less than AB the greater of 
the two sides AB, DC ; the straight line KA may, in the same 
manner, be demonstrated to be greater than the straight line 
drawn from the centre to the circumference of the circle 
described about the triangle. 

T3 



278 THE ELEMENTS 

PROP. XVII. PROB. 

seeN. I o describe in the greater of two spheres wliicli 
have the same centre, a solid polyhedron, the su- 
perficies of which shall not meet the lesser sphere. 

Let there be two spheres about the same centre A ; it is 
required to describe in the greater a solid polyhedron, the 
superficies of which shall not meet the lesser sphere. 

Let the spheres be cut by a planepassing through the centre ; 
the common sections of it with the spheres shall be circles ; 
because the sphere is described by the revolution of a semicir- 
cle about the diameter remaining unmoveable ; so that in what- 
, ever position the semicircle be conceived, the common section 

of the plane in which it is with the superficies of the sphere is 
the circumference of a circle : and this is a great circle of the 
sphere, because the diameter of the sphere, which is likewise 
» j5 3^ the diameter of the circle, is greater * than any straight line 
in the circle or sphere: Let then the circle made by the section 
of the plane with the greater sphere be BCDE, and with the 
lesser sphere be FGH ; and draw the two diameters BD, CE, 
at right angles to one another ; and in BCDE, the greater of 
j^ the two circles, describe '' a polygon of an even number of 

* ' equalsidesnotmeetingthelesser circle FGH 5 and let its sides, 
in BE the fourth part of the circle, be BK, KL, LM, ME ; 
join KAj and produce it to N ; and from A draw AX at right 
angles to the plane of the circle BCDE, meeting the super- 
ficies of the sphere in the point X j and let planes pass through 
AX, and each of the straight lines BD, KN, which, from 
what has been said, shall producegreat circles on the superficies 
of the sphere", and let BXD, KXN be the semicircles thus 
made upon the diameters BD, KN: Therefore, because XA 
is at right angles to the plane of the circle BCDE, every plane 
e 18. 11, which passes through X A is at right*^ angles to the plane of the 
circle BCDE ; wherefore the semicircles BXD, KXN are at 
right angles to that plane : Anil because the semicircles BED, 
BXD, KXN, upon the equal diameters BD, KN, are equal 
to one another, their halves BE, BX, KX, are equal to one 
another : Therefore, as many sides of the polygon as are in 
BE, so many there are in BX, KX equal to the sides BK, 
KL, LM, ME : Let these polygons be described, and their 
sides be BO, OP, PR, RX s KS, ST, l^Y, YX, and join 

OS, 



OF EUCLID. 



279 



OS, PT, RY ; and from the points O, S draw O V, SQ_per. Book XJI. 
pendiculars to AB, AK : And because the plane BOXD is at "■^^^^*^ 
right angles to the plane BCDE, and in one of them BOXD, 
O V is drawn perpendicular to AB the common seftion of the 
planes, therefore OV is perpendicular* to the plane BCDE 
For the same reason J^Q_ is perpendicular to the same plane, 
because the plane KSXN is at right angles to the plane BCDE. 
Join VQ; and because in the equal semicircles BXD, KXN 



»4.def. 11. 




the circumferences, BO, KS are equal, and OV, SQjare per- 
pendicular to their diameters, therefore'' OV is equal to SQj ■* 2i>. - 
and B V equal to KQ^ But the whole B A is equal to the whole 
KA, therefore the remainder VA is equal to the remainder 
QA : As therefore B V is to VA, so is KQ^to QA> wherefore 
VQ^is parallel' to BK: And because OV^, SQ_ are each of '2. 6. 
them at right angles to the plane of the circle BCDE, OV is 
parallel^ to SQ_; and it has been proved, that it is also equal '^6. 11 
to it ; therefore QV, SO are equal and parallel 8 : And because ' as. 1 
Oy is parallel to SO, and also to KB ; OS is parallel'' to BK ; » 9. n 
and therefore BO, KS which join them are in the same plane 

. T 4 in 



iSo 



THE ELEMENTS 



SooK. XII, in which these parallel are, and the quadrilateral figure KBOS 

^ is in one plane : And if BF, TK be joined, and perpendiculars 

be drawn from the points P, T to the straight lines AB, AK, 

it may be demonstras d, that TP is parallel to KB in the very 

same way that SO was shewn to be parallel to thcsameKBj 

»>. 11. wher fore* TP is parallel to SO, and the quadrilateral figure 
SOPT is in one plane; For the same reason the quadrilateral 
TPRY is in one plane : and the figure YRX is also in one 



"i. 11. 




planch Therefore, if from the points O, S, P, T, R, Y, there 
be drawn straight lines to the point A, there shall be formed 
a solid polyhedron between the circumferences BX, KX, com- 
posed of p) ram ids, the bases of which are the quadrilaterals 
KBOS, SOPT, TPRY, and the triangle YRX,and of which 
the common vertex is the point A : And if the same ponstruc- 
tion be made upon each of the sides KL, LM, ME, as has been 
done upon BK, and the like be done also in the other three 
quadrants, and in the other hemisphere ; there shall be 
formed a solid polyhedron described in the sphere, composed 
«f pyramids, the bases of which arc the aforesaid quadri- 
lateral 



OF EUCLID. 2S1 

lateral figures, and the triangle YRX, and those formed in BookXu. 
the lika manner in the rest ot the sphere, the common vertex 
of them all being the point A: and the superficies of this 
solid polyhedron does not meet the lesser sphere in which is the 
circle FGH : For, from the point A draw* AZ perpendicular * n. n, 
to the plane of the quadrilateral KBOS, meeting it in Z, and 
join BZ, ZK : And because AZ is perp;:-ndicular to the plane 
KBOS, it makes right angles with every straight line meeting 
it in that plane; therefore AZ is perpendicular to BZ and ZK: 
And because AB is equal to AK, and that tne squares of AZ, 
ZB, are equal to the square of AB ; and the squares of AZ, 
ZK to the square of AK''; therefore the sqtiares of AZ, ZB *47. i. 
are equal to the squares of AZ, ZK : Take from these equals 
the square of AZ, the remaining square of BZ is equal to the 
remaining square of ZK ; and therefore the straight line BZ 
is equal to ZK: In the like manner it may be demonstrated, 
that the straight lines drawn from the point Z to the points O, 
S are equal to BZ or ZK : Therefore the circle described from 
the centre Z. and distance ZB, shall passthrough the points K, 
O, S, and KBOS shall be a quadrilateral figure in th? circle: 
And because KB is greater than QV, and QV equal to SO, 
therefore KB is greater than SO : But KB is equal to each ot 
thes£raightlinesBO,KS; whereforeeach of the circumferences 
cut oft by KB, BO, KS is greater than that cut off^by OS; and 
these three circumferences, together with a fourth equal to one 
of them, are greater than the same three together with that cut 
ofl^ by OS ; that is, than the whole circumference of the cir- 
cle ; therefore the circumference subtended by KB is greater 
than the fourth part of the whole circumference of the circle 
KBOS, and consequently the angle BZK at the centre is 
greater than a right angle: And because the angle BZK is 
obtuse, thesquare of BK is greater'than the squares of BZ,ZK;' 12. 0. 
that is, greater than twice the square of BZ. Join KV, and 
becaus inthe triangles KBV,OBV, KB,BVareequaltoOB, 
BV, and that they contain equal angles ; the angle KBV is 
equal"" to the angle OVB : And OVB is a right angle ; there- 4 4, 1. 
fore also KV'B is a rioht angle : And because BD is less than 
twice DV; the rectangle contained by DB, BV is less than 
twice the rectangle Dv B ; that is% the square of KB is less* 8. 6. 
than twice the square of KV : But the square of KB is greater 
than twice the square of BZ ; therefore the square of KV is 
greater than the square of BZ : And because BA is equal to 
AK, and that the squares ol BZ, Z A are equal together to the 
square of BA, and the squares of KV, V A to the square of 

AK i 



282 THE ELEMENTS ^ 

Book xn. ^f^ . therefore the squares of BZ, ZA are equal to the square* 
^^'^■'"'^ of K V, VA ; and of tl>ese the square of K V is greater than the 
square of BZ ; therefore the square of VA is less than the 
square of ZA, and the straight line AZ greater than VA : 
Much more then is AZ greater than AG ; because, in the pre- 
ceding proposition, it was shewn that KV falls without the 
circle FGH : And AZ is perpendicular to the plane KBOS, 
and is therefore the shortest of all the straight lines that can be 
drawn from A, the centre of the sphere to that plane. There- 
fore the plane KBOS does not meet the lesser sphere. 

And that the other planes between the quadrants BX, KX 
fall without the lesser sphere, is thus demonstrated : From the 
point A draw AI perpendicular to the plane of the quadri- 
lateral SOPT, and join lO; and, as was demonstrated of the 
plane KBOS and the point Z, in the same way it may be shewn 
that the point I is the centre of a circle described about SOPT : 
and that OS is greater than PT; and PT was shewn to be 
parallel to OS : Therefore because the two trapeziums KBOS, 
SOFT inscribed in circles have their sides BK, OS parallel, as 
also OS, PT; and their other sides BO, KS, OP, ST, all equal 
to one another, and that BK is greater than OS, and OS 
• 2.Lem. greater than PT, therefore the straight line ZB is greater' 
than lO. Join AO which will be equal to AB ; and because 
AIO, AZB are right angles, the squares of AI, lO are equal 
to the square of AO or of AB i that is, to the squares of AZ, 
ZB ; and the square of ZB is greater than the square of lO, 
therefore the square of AZ is less than the square of AI ; and 
the straight line AZ less than the straight line AI : And it was 
proved, that AZ is greater than AG ; much more then is AI 
greater than AG : Therefore the plane SOPT falls wholly 
without the lesser sphere : In the same manner it may be de- 
monstrated, that theplaneTPRY falls withoutthe same sphere, 
as also the triangle YRX, viz. by the Cor. of 2d Lemma. And 
after the same way it may be demonstrated, that all the planes 
which contain the solid polyhedron, fall without the lesser 
sphere. Therefore ir>the greater of two spheres, which have 
the same centre, a solid polyhedron is described, the super- 
fices of which does not meet the lesser sphere. Which was 
to be done. 

But the straight line AZ may be demonstrated to be greater 
than AG otherwise, and in a shorter manner, without the help 
of Prop. 16, as follows. From the point G draw GU at right 
angles to AG, and join AU. -If then the circumferences BE bo 
bise<^d,-and its half again bisefted, and so on, there will at 

length 



OF EUCLID. 283 

Icngtfibcleftacircumferencelessthaathecircumference which B^jkXII. 
is subtended by a straight line equal to GU, inscribed in the *"^^ 
circle BCDE: Let this be the circuraference KB: Therefore 
the straight line KB is less than GU: And because the angle 
BZK is obtuse, as was proved in the preceding, therefore BK 
is greater than BZ : But GU is greater than BK ; much more 
then is GU greater than BZ, and the square of GU than tbe 
square of BZ j and AU is equal to AB ; therefore the square 
of AU, that is, the squares of AG, GU, are equal to the square 
of AB, that is, to the squares of AZ, ZB ; but the square of 
BZ is less than the square ofGU ; therefore the square of AZ 
is greater than the square of AG, and the straight line AZ 
consequently greater than the straight line AG. 

Cor. And if in the lesser sphere there be described a solid 
polyhedron, by drawing straight lines betwixt the points in 
which the straight lines from the centre of the sphere drawn 
to all the angles of the solid polyhedron in the greater sphere 
meet the superficies of the lesser ; in the same order in which 
are joined the points in which the same lines from the centre 
meet the superficies of the greater sphere ; the solid polyhe- 
dron in the sphere BCDE has to this other solid polyhedron 
the triplicate ratio of that which the diameter of the sphere 
BCDE has to the diameter »f the other sphere : For if these 
two solids be divided into the same number of pyramids, and 
in the same order, the pyramids shall be similar to one another, 
each to each : Because they have the solid angles at their 
common vertex, the centre of the sphere, the same in each 
pyramid, and their other solid angle at the bases equal to one 
another, each to each% because they are contained by three* B.ii. 
plane angles, each equal to each; and the pyramids are contained 
by the samenumber of similar planes; and are therefore similar'* "^^'^t^-^^* 
to one another, each to each : But similar pyramids have to 
one another the triplicate*^" ratio of their homologous sides. 'Cor.g.ia. 
Therefore the pyramid of which the base is the quadrilateral 
KBOS, ami vertex A, has to the pyramid in the other sphere 
of the same order, the triplicate ratio of their homologous 
sides, that is, of that ratio which AB from the centre of the 
greater sphere has to the straight line from the same centre to 
tae superficies of the lesser sphere. And in like manner, each 
pyramid in the greater sphere has to each of the same order in 
the lesser, the triplicate ratio of that which AB has to the 
semidiameter of the lesser sphere. And as one antecedent is to 
its consequent, so are all the antecedents to all the consequents. 
Wherefore the whole solid polyhedron in the greater sphere has 
to the whole solid polyhedron in the other, the triplicate ratio 

of 



he 



284 THE ELEMENTS 

nc 

B(x>K XII. of that which AB the semidiameter of the first has to t q^^- 

^^''^"'^ midiameter of the other ; that is, which the diameter B 0^ 
the greater has to the diameter of the other sphere. 



» n. 12. 



PROP. XVIII. THEOR. 

oPHERES have to one anotlier the triplicate ratio 
of that which their diameters have. 

Let ABC, DEF be two spheres, of which the diameters 
are BC, EF. The sphere ABC has to the sphere DEF the 
tripHcate ratio of that which BC has to EF. 

For, if it has not, the sphere ABC shall have to a sphere 
either less ®r greater, than DEF, the triplicate ratio of that 
which BC has to EF. First, let it have that ratio to a less, viz. 
to the sphere GHK ; and let the spliere DEF have the same 
centre with GHK j and in the greater sphere DEF describe" 




a solid polyhedron, the superficies of which does not meet the 
lesser sphere GHK ; and in the sphere ABC describe another 
similar to that in the sphere DEF : Therefore the solid poly- 
hedron in the sphere ABC has to the solid polyhedron in the 

Cor. 17. sphere DEF, the triplicate ratio^ of that which BC has to EF. 
But the sphere ABC has to the sphere GHK, the triplicate 
ratio of that which BChas toEF; therefore as the sphere ABC 
to the sphere GH K, so is the said polyhedron in ihcsphere ABC 
to the solid polyhedron in the sphere DEF : But the sphere 

= 14.5. ABC is greater than the solid polyhedron in it; therefore* 
also the sphere GHK is greater than thesolid polyhedron in the 
sphere DEF : But it is also less, because it is contained within 
it, which is impossible ; Therefore the sphere ABC has not to 

any 



OF KUCLID. 285 

any sphere less than DEF, the triplicate ratio of that which ^<^^ xir. 
BC has to EF. In the same manner, it may be demonstrated, '"^^^'^^ 
that the sphere DEF has not to any sphere less than ABC, the 
triplicate ratio of that which EF has to BC. Nor can the 
sphere ABC have to any sphere greater than DEF, the tripli- 
cate ratio of that which Be has to EF : For, if it can, Jet it 
have that ratio to a greater sphere LiVlN : Therefore, by in- 
version, the sphere LMN has to the sphere ABC, the tripli- 
cate ratio of that which the diameter EF has to the diameter 
BC. But as the sphere LMN to ABC, so is the sphere DEF 
to some sphere, which must be less'^ than the sphere ABC, 
because the sphere LMN is greater than the sphere DEF : 
therefore the sphere DEF has to a sphere less than ABC the 
triplicate ratio of that which EFhas toBC ; which was shewn 
to be impossible ; Therefore the sphere ABC has not to any 
sphere greater than DEF the triplicate ratio of that which 
BC has to EF : and it was demonstrated, that neither has it 
that ratio to any sphere less than DEF. Tnerefore the sphere 
ABC has to the sphere DEF, the triplicate ratio of that which 
BC has to EF. Q. E. D. 



END OF THE ELEMENTS. 



I 



NOTES, 



CRITICAL AND GEOMETRICAL ; 



CONTAINING 



An Account of those Things in which this Edition 
differs from the Greek Text ; and the Reasons of 
the Alterations which have been made. As also 
Observations on some of the Propositions. 



By ROBERT SIMSON, M. D. 

Emeritus Professor of Mathematics in the University of Glasgow, 



LONDON: 

Printed for F. Wingrave, in the Strand, Successor to Mr. Noubse, 
1806. 



NOTES, &c. 



aS9 



DEFINITION I. BOOK I. 

It is necessary to consider a solid, thatis, a magnitude which Book i. 
has length, breadth, and thickness, in order to understand *-^v"*' 
aright the definitions of a point, line, and superfices ; for these 
all arise from a solid, and exist in it : The boundary, or boun- 
daries which contain a solid are called superficies, or the boun- 
dary which is common to two solids which are contiguous, or 
which divides one solid into two contiguous parts, is called a 
superficies : Thus, if BCGF be one of the boundaries which 
contain the solid ABCDEFGH, or which is the common 
boundaryofth.ssoiid,andtheso|id BKLCFNMG,and isthere- 
forein the one as well as in the other solid, is called a superficies, 
and has no thickness : For, if it have any, this thickness must 
either be; a part of the thickness 



H G 



,./ k/ 


NT 


/ 


B 




C 




x/ 


/ 


/ 



M 



.\ 



B 



of the solid AG, or of the solid 
BM, or a part of the thickness 
of each of them. It cannot be a 
part of the thickness of the solid 
BM ; because if this solid be 
removed from the solid AG, the 
superficies BCGF, the boundary 
of the solid AG, remain still the 
same as it was. Nor can it be a 

part of the thickness of the solid AG j because if this be re- 
moved from the solid BM, the superficies BCGF, the boundary 
of the solid BM does nevertheless remain, therefore the super- 
fices BCGF has no thickness, but only length and breadth. 

The boundary of a superficies is called a line, or a line is the 
common boundary of two superficies that are contiguous, or 
which divides one superficies into two contiguous parts : Thus 
if BC be one of the boundaries which contain the superficies 
ABCD, or which is the common boundary of this superficies, 
and of the superficies KBCL which is contiguous to it, this 
boundary BC is called a line, and has no breadth : For if it 
have any, this must be part either of the breadth of the super- 
ficies ABCD, or of the' superfices KBCL, or part of each of 
them. It is not part of the breadth of the superfices KBCL ; 
for, if this superfices be removed from the superfices ABCD, 

U th(; 



290 

Book I. 



■/ 


f/ 


V 


/ 


T> 






C 




/ 




/ 


/ 


A. 


B 


K 



NOTES. 

the line BC, which is the boundary of the supeifices ABCD, 
remains the same as it was: Nor can the breadth that BC is 
supposed to have, be a part of the breadth of the superficies 
ABCD J because, if this be removed from the superficies 
KBCL, the line BC, which is the boundary of the superficies 
KBCL, d#>es nevertheless remain : Therefore the line BC has 
no breadth : And because the line BC is in a superficies, and 
that a superficies has no thickness, as was shewn, therefore a 
line has raither breadth nor thickness, but only length. 

Tlie boundary of a line is called a point, or a point is the 
common boundary or extremity 

of two lines that are contiguous : 1\ G 1^1 

Thus, if B be the extremity of 
the line AB, or the commoii ex- 
tremity of the two lines AB, KB, 
this extremity is called a point, 
and has no length : For if it have 
any, this length must either be 
part of the length of the line AB, 
or of the line KB. It is not part 
of the length of KB ; for if the line KB be removed from AB, 
the po1nt"^B which is the extremity of the line AB remains the 
same as it was : Nor is it part of the length of the line AB ; for, 
if AB be removed from the line KB, the point B, which is the 
extremity of the line KB, does nevertheless remain : rherefore 
the point B has nolength : And because a point js in a line, and 
a line has neither breadth nor thickness, therefore a point has 
no length, breadth, nor thickness. And in this manner the defi- 
nitions of a point, line, and superficies, are to be understood. 

DEF. VII. B. I. 

Instead of this definition as it is in the Greek copies, a 
more distinct one is given from a property of a plane super- 
ficies, which is manifestly supposed in thelElements, viz. that 
a st/aight line drawn from any point in a plane to any other 
in it, is wholly in that plane. 

DEF. VIII. B. I. 

It seems that he who made this definition designed that it 
should comprehend not only a plane angle contatned by two 
straight lines, but likewise the angle which some conceive to 
be made by a straight Ime and a curve, or by two curve lines ( 
which meet one another in a plane : But, though the meaning of 

the 



NOTES. 291 

the words c-n Ey^£»2f, that is, in a straight line, or in the same BjokL 
direction, be plain, when two straight lines are said to be in a ^"^' 
straight line, it does not appear what ought to be understood 
by these words, when a straight line and a curve, or two curve 
lines, are said to be in the same direction ; at least it cannot be 
explained in this place ; which makes it probable that this de- 
iinition, and that of the angle of a segment, and what is said 
of the angle of a semicircle, and the angles of segments, in the 
i6thand 31st Propositions of Book 3, are the additions of some 
less skilful editor : On which account, especially since thejr 
are quite useless, these definitions are distinguished from the 
rest by inverted double commas. 

DEF. X\ai. B. I. ■ 

The words, " which also divides the circle into two equal 
*' parts" are added at the end of this definition in all the copies, 
but are now Irft out as not belonging to the definition, being 
only a corollory from ic. Procius demonstrates it by conceiv- 
ing one of the parts into which the diameter divides the circle, 
to be applied to the other j for it is plain they must coincide, 
else the straight lines from the centre, to the circumference 
would not be all equal : The same thing is easily deduced from 
the 31st Prop, of Book 3, and the 24ch of the same; from the 
first of which it follows, that semicircles are similar segments of 
a Circle ; and from the other, that they are equal to one another. 

DEF. XXXIII. B. I. 

This definition has one condition more than is necessary; 
because every quadrilateral figure which has its opposite sides 
equal to one another, has likewise its opposite angles equal ; 
and on the contrary. 

Let ABCD be a quadrilateral figure, of which the opposite 
sides AB, CD, are equal to one ano- 
ther ; as also AD and BC : Join BD i 
the two sides AD, DB are equal to 
the two CB, BD, and the base AB is 
equal to the base CD ; therefore, by 
Prop. 8. of Book I. the angle ADB is 

.equal to the an^le CBD; and, by Prop. 4. B. i. the angle 
BAD is equal ^to the angle DCB, and ABD to BDC ; and 
therefore also the angle ADC is equal to the angle A^C. 

U X And 





292 N O T E S. 

^ooK^ And if the angle BAD be equal to the opposite angle BCD, 
'*^^^^''*^ and the angle ABC to ADC; the opposite sides are equal: 
Because, by Prop. 32. B: i. all the angles of the quadrilateral 
figure ABCD are together equal to 
four right angles, and the two angles 
BAD, ADC are together equal to 
the two angles BCD, ABC : Where- 
fore BAD, ADC are the half of all 
the four angles ; that is, BAD and " 
ADC are equal to two right angles : and therefore AP, CD 
are parallels by Prop. 28. B. i. In the same manner, AD, BC 
are parallels : Therefore ABCD is a parallelogram, and its 
opposite sides are equal, by 34th Prop, B. 1. 

PROP. VII. B. I. 

There are two cases of this proposition, one of which is 
not in the Greek text, but is as necessary as the other : And 
that the case left out has been formerly in the text, appears ■ 
plainly from this, that the second part of Prop. 5. which is 
necessary to the demonstration of this case, can be of no use at 
all in the Elements, or any where else, but in this demonstra- 
tion ; because the second part of Prop. 5. clearly follows from 
the first part, and Prop. 13. B. i. This part must therefore 
have been added to Prop. 5, upon account of some proposition 
betwixt the 5th and 13th, but none of these stand in need of 
it except the 7 th Proposition, on account of which it has been 
added : Besides, the translation from the Arabic has this case 
explicitly demonstrated. And Proclus acknowledges, that the 
second Part of Prop. 5. was added upon account of Prop. 7. 
but gives a ridiculous reason for it, " that it might afford an 
" answer to objections made against the 7th," as if the case of 
the 7th, which is left out, were, as he expressly makes it, an 
objection against the proposition itself- Whoever is curious 
may read what Proclus says of this in his commentary on the 
5th and 7th Propositions ; for it is not worth while to relate 
his trifles at full length. , ► . 

It was thought proper to change the enunciation of this jth 
Prop, so as to preserve the very same meaning ; the literal 
translation from the Greek being extremely harsh, and diffi- 
cult to be understoud by beginners. 



NOTES. 

PROP. XL B. I. 

A COROLLARY IS added to this proposition, which is ne- 
cessary to Prop. I. B. II. and otherwise. 

PROP. XX. and XXI. B. I. 

Proclus, in his commentary, relates, that the Epicureans 
derided this proposition, as being manifest even to asses, zni 
needing no demonstration; and his answer is, that though the 
truth of it be manifest to our senses, yet it is science which 
must give the reason why two sides of a triangle are greater 
than the third : But the right answer to this obje<Slion against 
this and the 21st, and some other plain propositions, is, that 
the number of axioms ought not to be increased without ne- 
cessity, as it must be if these propositions be not demonstrated. 
Mons. Clairauit, in the Preface to his Elements of Geometry, 
published in French at Paris, anno 1741, says. That Euclid 
has been at the pains to prove, that the two sides of a triangle 
which is included within another, are together less than the 
two sides of the triangle which includes it ; but he has forgot 
to add this condition, viz. that the triangks must be upon the 
same base ; because, unless this be added, the sides of the in- 
cluded triangle may be greater than the sides of the triangle 
which includes it, in any ratio which is less than that of two 
to one, as Pappus Alexandrinus has demonstrated in Prop. 3. 
B. 3. of his mathematical collections. 



293 



Book I. 



PROP. XXII. B. I. 

Some authors blame Euclid because he does not demonstrate 
ir.it the two circles made use of in the construction of this 
problem must cut one another : But this is very plain from the 
determination he has given, viz. that any two of the straight 
lines DF, FG, GH must be 
greater than the third : For who 
is so dull, though only beginning 
to learn the Elements, as not to 
perceive that the circle describ- 
ed from the centre F, at the 
distance FD, must meet FH 
betwixt F and H, because FD 

is less than FH ; and that, , for the like reason, the circle de- 
scribed from the centre G, at the distance GH or GM, must 

U 3 meet 




294 



NOTES. 




,^°° K I. meet DG betwixt D and G ; and that these circles must meet 
^ one another, because FD and 
GH are together greater than 
FG? And this determination is 
easier to be understood than that 
which Mr. Thomas Simpson 

derives from it, and puts instead __^ 

of Euclid's, in the 49th page of ]5~M Y G H 

his Elements of Geometry, that 

he may supply the omission he blames Euclid for, which de- 
termination is, that any of the three straight lines must be less 
than the sum, but greater than the difference of the other two: 
From this he shews the circles must meet one another, in one 
case ; and says, that it may be proved after the same manner in 
any other case ; But the straight line GiVI, v/hich he bids take 
■^ from GF may be greater than it, as in the figure here annexed ; 

in which case his dem.onstration must be changed into another. 



PROP. XXIV. B. I. 

To this is added, " of the two sides DE, DF, let DE be 
** that which is not greater than the other j" that is, take that 
side of the twoDE, DF which is not greater than the other, in 
order to make with It the angle EDG jy 
equal to BAG; because without this 
restriction there mig-ht be three diffe- 
rent cases of the proposition, as Cam- 
panus and others make. 

Mr. Thomas Simpson, in p. 262 of 
the second edition of his Elements of 
Geometry, printed anno 1760, ob- 
serves in his notes, that it ought to 
have been shewn, that the points F 
fall below the line EG. This proba- 
bly Euclid omitted, as it is very easy to perceive, that DG 
being equal to DF, the point G is in the circumference of a 
circle described from the centre D at the distance DF,and must 
be in that part of it which is above the straight line EF, be- 
cause DG falls above DF, the angle EDG being greater than 
the angle EDF. 

PROP. XXIX. B. I. 

The propositiort which isusuallycalled the 5th postulate, or 
jith axiom, by some the i2th,on which this 29th depends, has 

given 




NOTES. 295 

g-iven a great deal to do, both to ancient and modern geometers : Boo k i. 
It seecns not to be properly placed among the a-xioms, as indeed '^^^•^^^ 
it is not self-evident ; but it may be demonstrated thus : 

DEFINITION I. 

The distance of a point from a straight line, is the perpen- 
dicular drawn to it from the point. 

DEF. 2. 

One straight line is said to go nearer to, or further from, 
another straight line, when the distances of the points of the 
first from the other straight line become less or greater than 
they were ; and two straight lines are said to keep the same 
distance from one another, when the distance of the points of 
one of them from the other is always the same. 

AXIOxM. 

A STRAIGHT line cannot first come nearer to another 
straight line, and then go further 
from it, before it cuts it; and, 
in like manner, a straight line 
cannot go further fr^m another 
straight line,and then come nearer 
to it; nor can a straight line keep 

the same distance from another straight line, and then come 
nearer to it, or go further from it; for a straight line keeps 
always the same direction. 

For example, the straight line ABC cannot first come nearer 
to the straight line DE,as from the 

point A to the point B, and then, ^ g See the 

from the point B to the point C, go ^') ^ above 

further from the same DE : And, in _ ____^ E ^s«fe- 

like manner, the straight line FGH ^ G H 

cannot go further from DE, as from 

F to G, and then, from G to H, come nearer to the same DE : 
And so in the last case, as in fig. 2. 

PROP. I. 

If two equal straight lines AC, BD, be each at right 
angles to the same straight line AB : If the points C, D be 
joined by the straight line CD, the straight line EF drawn 
from any point E in AB unto CD, at right angles to AB,'shall 
be equal to AC, or BD. 

If EF be not equal to AC, one of them must be greater 
than the other ; let AC be the greater j then, because FE is 

U 4 less 




296 



NOTES. 



= 4. 1. 



s. 1. 




BookJL^ less than CA, the straight line CFD is nearer to the straight 
line AB at the point F than at the 
point C, that is, CF comes nearer 
to AB from the point C to F : But 
because DB is greater than FE, 
the straight line CFD is further 
from AB at the point D than at F, 
that is, FD goes further from AB 
from F to D : Therefore the 
straight line CFD first comes 
nearer to the straight line AB, and then goes further from it, 
before it cuts it ; which is impossible. Jf FE be said to be 
greater than CA, or DS, the straight line CFD first goes fur- 
ther from the straight line AB, and then comes nearer to it: 
which is also impossible. Therefore FE is not unequal to AC, 
that is, it is equal to it. 

PROP. II. 

If two equal straight lines AC, BD be each at right angles 
to the same straight lineAB; the straight line CD which 
joins their extremities makes right angles with AC and BD. 

Join AD, BCj and because, in the triangles CAB, DBA, 
CA, AB are equal to DB, BA, and the angle CAB equal to 
the angle DBA ; the base BC is equal » to the base AD : And 
in the triangles ACD, BDC, AC, CD are equal to BD, DC, 
and the base AD is equal to the base 
BC : Therefore the angle ACD is 
equaP to the angle BDC : From 
any point E in AB draw EF unto 
CD, at right angles to AB ; there- 
fore, by Prop. I. EF is equal to AC, 
or BD ; wherefore, as has been just 
now shewn, the angle, ACF is equal 
to the angle EFC : In the same manner 




IS 



the angle BDF 
equal to the angle EFD ; but the angles ACD, BDC are equal ; 
10. def. 1. therefore the angles EFC and EFD arc equal, and right angles^ j 
wherefore also the angles ACD, BDC are right angles. 

CoR. Hence, if two straight lines AB, CD be at right an- 
gles to the same straight line AC, and if betwixt them a 
straight line BD be drawn at right angles to cither of them, as 
to AB; then BD is equal to AC, and BDC is a right angle. 

If AC be not equal to BD, take BG equal to AC, and join 
CG: Therefore, by this proposition, the angle ACG is a right 
anglpi but ACD is also a right angle j wherefore the angles 

ACD, 



297 

Book I. 



N O T £ S. 

ACD, ACG are equal to one another, whnch is impassible. 
Therefore BD is equal to AC j and by this proposition BDC "'^^'" 
is a right angle. 

PROP. 3. 

If two straight lines which contain an angle be produced, 
there may be found in either of them a point from which the 
perpendicular drawn to the other shall be greater than any 
given straight line. 

Let AB, AC be two str^ght lines which make an angle with 
one another, and let AD hi the given straight line; a point 
may be found either in ABor AC, as in AC, from which the 
perpendicular drawn to the other AB shall be greater than AD. 

In AC take any point E, and draw EF perpendicular to 
AB ; produce AE to G, so that EG be equal to AE j and 
produce FE to H, and make EH equal to FE, and join HG. 
Because, in the triangles AEF, GEH, AE, E!F are equal to 
GE, EH, each to each, and contain equal * angles, the angle » 15 j_ 
GHE is therefore equal '' to the an::le AFE which is a right ^^ ( 
angle : Draw GK perpendicular to AB ; and because the straight 
lines FK, HG 
are at right an- ^ ^^ t -d ,- 

glestoFH, and .p ^'- ^ ^ ^^^ 

KG at right an- "■'" *- 

gles to FK, KG 

is equal to FH, 

by Cor. Pr. 2. 

that is, to the 

double of FE. 

In the same manner if AG be produced to L, so that GL be 

equal to AG, and LM be drawn perpendicular to AB, then 

LM is double of GK, and so on. In AD take A\ equal to 

FE, and AO, equal to KG, that is, to the double ofFE, or 

AN; also, take AP, equal to LM, that is, to the double of 

KG, or AO ; and let this be done till the straight line taken 

be greater than AD : Let this straight line so taken be AP, 

and because AP is equal to LM, therefore LM is greater than 

AD. Which was to be done. 




PROP. 4. 

If two straight lines AB, CD make equal angles EAB, 
ECD with another straight line EAC towards the same parts 
of it; AB and CD are at right angles to some straight line. 

Bisect 



Book I. 



» 15. I. 

"4. 1. 



298 N O T E S. ^ 

Biseel: AC in F, and draw ¥G perpendicular to AB ; take 
CH in the straight ]ine CD, equal to AG, and on the contrary 
side of AC to that on which AG is, and join FH : Theretbre, 
in the triangles AFG, CFH, the sides FA, AG are equal to 
FC, CH, each to each, and the angle 
FAG, that^ is EAB, is equal to the 
an^le FCH ; wherefore*' the angle 
AGF is equal to CHF, aiui AFG to 
the angle CFH: To these last add the 
common angle AFH -, therefore the 
two angles AF"G, AFH are equal- 
to the two angles CFH, HFA, 
which two last are equal together to 

• \^. i. two right angles*^ : therefore also 

* li. 1, AFG, AFH are equal to two right angles, and consequently** 

GF and FH are in one straight line. And because AGF is a 
right angle, CHF which is equal to it is also a right angle ; 
Therefore the straight lines AB, CD are at rightangles to GH. 





C H 



I) 



2C\ I. 



•> 13. 1. 



PROP. 5. 

If two straight lines BB, CD be cut by a third ACE, so as 
to make the interior angles BAC, ACD, on the same side of 
it, together less than two right angles; AB and CD being 
produced, shall meet one another towards the parts on which 
are the two angles, which are less than two right angles. 

At the point C, in the straight line CF, make^ the angle 
ECF equal to the angle EAB, and draw to AB the straight 
line CG at right angles to CF : Fhen, because the angles ECF, 
E AB^are equal to one an- 
other, and that the angles K 
ECF, B'CA are together 
equal*" to two right an- 
gle's, the angles EAB, 
FCA are equal to two 
right angles. But by the 
hypothesis, the angles 
EAB, ACD are toge- 
ther less than two right 
angles ; therefore theangle 

FCA is greater than ACD, and CD falls between CF and 
AB : And because CF and CD make an angle with one ano- 
tber, by Prop. 3. a point may be found in either of them CD^ 
from which the perpendic-ular drawn to CF shall be greatCj. 




N O T E S. 299 

than tlie sfraight line CG. Let thrs point be H, and dtaw ,^^^^J\ 
HK perpendicular to CF, meeting AB in L : And because 
AB, CF contain equal angles with AC on the same side of it, 
by Prop. 4. AB and CK are at right angles to the straight 
line MNO, which bisects AC in N,and is perpendicular to CF ; 
Therefore by'Cor. Prop. 2. CGand KL, which are at right 
angles to CF, are equal to one another : And HK is greater 
than CG, and therefore is greater than KL, and consequently 
the point H is in KL produced. Wherefore the straight line 
CDH, drawn betwixt the points C, H, which are on contrary 
sides of AL, must necessarily cut the straight line AB. 

PROP. XXXV. B. T. 

The demonstration of this Proposition is changed, because, 
if the method which is used in it was followed, thtrre would be 
three cases to be separately demonstrated, as is done in the 
translation from the Arabic; for, in the Elements, no case of 
a Proposition that requires a different demonstration ought"tp 
be omitted. On this account, we have chosen the method 
which Morfs. Ciairault has given, the first of any, as far as I 
know, in his Elements, page 21, and which afterwards Mr. 
Simpson gives in his page 32. But whereas Mr. Simpson 
makes use of Prop. 26. B. i. from which the eq'jality of the 
two triangles does not immediately follow, because, to prove 
that, the 4th of B. 1. must likewise be»made use of, as may be 
seen in the very same case in the 34th Prop. B. i. it was 
thought better to make use only of the 4th of B. i. 

PROP. XLV. B.I. 

The straight line KM is proved to be parallel to FI, from 
the 33d Prop, whereas KH is parallel to FGby contruction, 
and KHAl, FGL have been demonstrated to be ^-.traight lines. 
A corollary is added from Commandine, as being often used. 

PROP. xiir. B. II. 

J.N this proposition only acute angled triangles are men- BookTL 
tioned, whereas it holds true of every triangle: and the demon- ^•^s'^^ 
strations of the cases omitted are added; Commandine andCla- 
vius have likewise given their demonstrations of these cases. 

PROP. XIV. B. II. 

In the demonstration of this, some Greek editor lias igno- 
rantly inserted the words, " but if not, one of the two BE, 

" ED, 




NOTES. 

" ED, is the greater t Let BE Ijc the greater, and produce it to 
" F," as if it was of any consequence whether the greater or 
lesser be produced : Therefore, instead of the:?e words, there 
ought to be read only, *' but if not, produce BE to F." 



Book III. Q 



PROP. I. B. Ill, 



EVERAL authors, especially among the modern mathe- 
maticians' and logicians, inveigh too severely against indirect 
or apogogic demonstrations, and sometimes ignorantly enough; 
not bemg aware that there are some things that cannot be de- 
monstrated any other way : Of this thu present proposition 
is a very clear instance, as no direct demonstration can be 
given of it : Because, besides the definition of a circle, there 
is no principle or property relating to a circle antecedent to 
this problem, from which either a direct or indirect demon- 
stration can be deduced : Wherefore it is necessary that the 
point found by the construction of the problem be proved to 
be the centre of the circle, by the help of this definition, and 
some of the preceding propositions : And because, in the de- 
monstration, this proposition must be brought in, viz. straight 
lines from the centre of a circle to the circumference are equal, 
and that the point found by the construction cannot be as- 
sumed as the centre, for. this is the thing to be demonstrated : 
it is manifest some other point must be assumed as the centre ; 
and if from this assumption an absurdity follows, as Euclid de- 
monstrates there must, then it is not true that the point as- 
sumed is the centre; and as any point whatever was assumed, 
it follows that no point, except that found by the construction, 
can be the centre, from which the necessity of an indirect 
demonstration in this case is evident. 



PROP. XIII. B. III. 

As it is much easier to imagine that two circles may touch 
one another within in more points than one, upon the same 
side, than upon opposite sides ; the figure of that case ought 
not to have been omitted ; but the construction in the Greek 
text would not have suited with this figure so well, because the 
centres of the circles must have been placed near to the cir- 
cumferences; on which account another construction and 
demonstration is given, which is the same with the second part 
of that which Campanus has translated from the Arabic, 

where, 



NOTES 3^1 

where, without any reason, the demonstration is divided into ^0°^^- 
two parts. 

PROP. XV. B. III. 
Th£ converse of the second part of this proposition is want- 
ing, though in the preceding, the converse is added, in a like 
case, both in the enunciation and demomtration; and it is now 
added in this. Besides, in the demonstration of the first part 
of this I5thj tiiC diameter AD (see Commandine's figure), is 
proved to be greater than the straight iine BC by means of 
another straight line MN ; whereas it maybe better done v/ith- 
out it : on which accounts we have given a different demon- 
stration, like to that which Euclid gives m the preceding 14th, 
and to that which Theodosius gives in Prop. 6. B. i. of his 
Spherics, in this very affair. 

PRO?. XVI. B. III. 
In this we have not followed the Greek r.or the Latin trans- 
lation literally, but have given what is plainly the meaning of 
the proposition, without mentioning the angle ofthe semicircle, 
or that which some call the cornicular angb, which they con- 
ceive to be made by the circumference and the straight line 
which is at right angles to the diameter, at its extremity ; 
which angles have furnished matter of great debate between 
some ofthe modem gcometrers, and given occasion of deducing 
strange consequences from them, which are quite avoided by 
the manner in which we have expressed the proposition. And 
in like manner, we have given the true meaning of prop. 31. 
B. 3. without mentioning the angles ofthe greater or lesser 
segments. These passages Vieta, with good reason, suspetfls 
to be adulterated in the 386th page of his Oper. Math. . 

PROP. XX. B. III. 

The first words of the second part of this demonstration, 
'' xixXaffS'a/ '8r,zja'>.L\" are wrong translated by Mr. Briggs and 
Dr. Gregory, " Rursusinclinetur;" for the translation ought to 
be " Rursus infleclatur," as Commandine has it : A straight 
line is said to be inflected either to a straight, or curve line, 
when a straight line is drawn to this line from a point, and 
from the point in which it meets it, a straight line making 
an angle with the former is drawn to another point, as is evi- 
dent from the qoth prop, of Euclid's Data : For thus the whole 
line betwixt the first and last points is infledtd or broken at 

the 



302 NOTE S. 

Book in. the point of infle£lion, where the two straight lines meet. And 
in the like sense two straight lines are said to be infledled from 
two points to a third point, when they make an angle at this 
point; as may be seen in the description given by Pappus 
Alexandrinus of Appollonius's Books de Locis planis, in the 
preface to his yth Book : We have made the expression fuller 
from the 90th Prop, ot the Data. ' 

PROP. XXr. B. III. 

There are two cases of this proposition, the second of 
which, viz. when the angles are in a segment not greater than 
a semicircle, is wanting in the Greek : And of this a more 
simple demonstration is given than that which is in Comman- 
dine, as being derived only from the first case, without the 
help of triangles. 

PROP. XXIIl. and XXIV. B. III. 

In proposition 24. it is demonstrated that the segment 
AEB must coincide with the segment CFD (see Comman- 
dine's figure,) and that it cannot fail otherwise, as CGD, so as 
to cut the other circle in a third point G, from this, that, if it 
did, a circle could cut another in more points than two: But 
this ought to have been proved to be impossible in the 23d 
prop, as well as that one of the segments cannot fall '^'ithiri 
the other. This part, then, is left out in the 24th, and put 
in its proper place, the 23rd proposition. 

PROP. XXV. B. III. 

This proposition is divided into three cases, of which two 
have the same construilion and demoiistration j therefore it is 
now divided only into two cases. 

PROP. XXXIII. B. III. 

This also in the Greek is divided into three cases, o» 
which, two, viz. one, in which the given angle is acute, and 
the other in which it is obtuse, have exactly the same con- 
strudlion and demonstration : on which account, the demon- 
stration of the last case is left out, as quite superfluous, and 
the addition of some unskilful editor j besides the demonstra- 
tion of the case when the angle given is a right angle, is done 
a roundirabout way, and is therefore changed to a more sim- 
ple one,- as was done by Clavius. 



NOTES. 

PROP. XXXV. B. 111. 
As the 25rh and 33'-<i propositions are divided into more 
oases, so this 35th is divided into fewer cases than are necessa- 
ry. Nof can It be supposed that Euclid omitted them because 
they are easy ; as he has given the case, which by far is the 
easiest of them all, viz. that in which both the straight lines 
pass through the centre: And in the following proposition he 
separately demonstrates the case in which the straight line 
passes through the centre, and that in which it does not pass 
through the centre ; So that it seems Theon, or some other, 
has thought them too long to insert : But cases that require 
different demonstrations, should not be leftout in the Elements, 
a^ was before taken notice of: These cases are in the transla- 
tion from the Arabic, and are now put into the text. 

PROP. XXXVII. B. III. 

At the end of this, the words " in the same manner it may 
" be demonstrated, if the centre be in AC," are left out as the 
addition of some ignorant editor. 




w, 



DEFINITIONS of BOOK IV. 



HEN a point is in a straight line, or any other line, this Book IV 
point is by the Greek geometers said arareffS-ai, to be upon, "-^^^"^ 
or in that line, and when a straight line or circle meets a cir- 
cle any way, the one is said u^rsa^ai to meet the other : But 
when a straight line or circle meets a circle so as not to cut it, 
it is said g^pasTTSffS-aj, to touch the circle ; aqd these two terms 
are never promiscuously used by them : Therefore, in the 5th 
definition of B. 4. the compound £{paiSTr/T«imustbe read, instead 
of the simple aSJTy/Tsci : And in the ist, 2d, 3d, and 6th defi- 
nitions in Commandine's translation, " tangit," must be read 
instead of" contingit :" And in the 2d and 3d definitions of 
Book 3. the same change must be made : But in the Greek 
text of propositions nth, 12th, 13th, i8th, 19th, Book 3. 
the compound verb is to be put for the simple. 

PROP. IV. B. IV. 

In this, as also in the 8th and I3ch proposition of this book, 
it is demonstrated indirectly, that the circle touches a straight 
line; whereas in the 17th, 33d, and 37th propositions of Book 
3. the same thing is directly demonsti'ated : And this way we 

have 



394 NOTES. 

Book IV. havc chosci) to usc in the propositions of this book, as it is 
""^^^'"^ shorter. 

PROP. V. B. IV. 

The demonstration of this has been spoiled by some un- 
skilful hand: For he does not demonstrate, as is necessary, 
that the two straight lines which bisedls the sides of the tri- 
angle at right angles must meet one another ; and, without any 
reason, he divides the proposition into three cases j whereas, 
one and the same construdfion and demonstration serves for 
them all, as Campanus has observed ; which useless repetitions 
are now left out : The Greek text also in the corollary is 
manifestly vitiated, where mention is made of a given angle, 
though there neither is, nor can be, any thing in the proposi- 
tion relating to a given angle. 

PROP. XV. and XVI. B. IV. 
In the corollary of the first of these, the words equilateral 
and equiangular are wanting in the Greek : and in prop. i6. 
instead of the circle ABCD, ought to be read the circumfe- 
rence ABCD : Where mention is made of its containing fif- 
teen equal parts. 



DEF. III. B. V. 

JooK V jLVi-ANY of the modern mathematicians rejeft this defini- 
'"^^'^'^ tion : The very learned Dr. Barrow has explained it at large 
at the end of his third lefture of the year 1666, in which also 
he answers the objections made against it as well as the sub- 
ject would allow : And at the end gives his opinion upon the 
whole, as follows ; 

" I shall only add, that the author had, perhaps, no other 
** design in making this definition, than (that he might more 
*' fully explain and embellish his subjecl) to give a general 
*' and summary idea of ratio to beginners, by premising 
" this metaphysical definition, to the more accurate defini- 
*' tioiis of ratios that are the same to one another, or one of 
*^ which is greater, or less than the other: I call it a meta- 
" physical, for it is not properly a mathematical, definition, 
*' since nothing in mathematics depends on it, or is deduced, 
*' nor, as I judge, can be deduced from it : And the defini- 
** tion of analogy, which follows, \\z. Analogy is the simi- 
3 • " litudc 



NOTES. 



3»5 



" litude of ratios, is of the same kind, and can serve for no B^ok v. 

*' purpose in mathematics, but only to give beginners some ge- ^"^""^^ 

** neral, though gross and confused, notion of analogy: But the 

** whole of the doilrine of ratios, and the whole of mathema- 

*' ticsjdependupontheaccurate mathematical definitions which 

*' tallow this : To these we ought principally to attend, as the 

** doctrine of ratios is more perfectly explained by them j this 

*' third, and others like it, may be entirely spared without any 

*' loss to geomecry ; as we see in the 7th book of the Elements, 

*' where the proportion of numbers to one another is defined, 

** and treated of, yet without giving any definition of the ratio 

" of numDers ; though such a definition was as necessary and 

*' useful to be given in that hookas in this : But indeed there 

*' is scarce any need of it in either of them : Though I think 

' ' that a thing of so general and abstracted a nature, and there- 

*' by the more difficult to be conceived and explained, cannot 

*' be more commodiously defined than as the author has done : 

*' Upon which account I thought fit to explain it at large, and 

" defend it against the captious objections of those who attack 

*' it." To this citation from Dr. Barrow I have nothing to 

add, except that I fully believe the 3d and 8th definitions are 

not Euclid' S) but added by some unskilful editor. 

DEF. XL B. V. 

It was necessary to add the word " continual" before 
" proportionals" in this definition; and thus it is cited in the 
33d prop, of Book II. 

After this definition ought to have followed the definition of 
compound ratio, as this was the proper place for it; duplicate 
and triplicate ratio being species of compound ratio: ButTheon 
has made it the 5th def. of B. 6. where he gives an absurd and 
entirely useless definition of compound ratio : For this reason 
we have placed another definition of it betwixt the nth and 
I2th of this book, which, no doubt, Euclid gave; for he cites 
it expressly in Prop. 23. B. b, and which Clavius, Herigon, 
and Barrow, have likewise given, but they retain also Theon's, 
which they ought to have left out of the Elements. 

DEF. XIII. B. V. 

This, and the rest of the definitions following, contain the 
explication of some terms which are used in the 5th and foj.. 
lowing books; which, except a icw, are easily enough under* 

X stood 



3o6 NOTES. 



Book V. 



Stood from the propositions of this book where they are first 
mentioned : They seem to have been added by I'heon, or 
some other. However it be, they are explained something 
more distinctly for the sake of learners. 

PROP, IV. B. V. 

In the construftion preceding the demonstration of this, 
the words » £Tyx,£) any wliatever, are twice wanting in the 
Greek, as also in the Latin translations j and are now added, 
as being wholly necessary. 

Ibid, in the demonstration ; in the Greek, and in the Latin 
translation of Commandine, and in that of Mr. Henry Briggs, 
which was published at London in 1620, together with the 
Greek text of the first six bocks, which translation in this place 
is fallowed by Dr. Gregory in his edition of Euclid, there is this 
sentence following, viz. " and of A and C have been taken 
." equimultiples K, L ; and of B and D, any equimultiples 
" whatever (« ct^xO ^i ^ ;" which is not true, the words 
" any whatever," ought to be left out : And it is Strange that 
neither Mr. Briggs, who did right to leave out these words in 
one place of Prop. 13, of this book, nor Dr. Gregory, who 
changed them into the word *•' some" in three places, and left 
them out in a fourth of that same Prop. 13, did not also leave 
them out in this place of Prop. 4, and in the second of the two 
places where they occur in Prop. 17 of this book, in neither of 
which they can stand consistent with truth : And in none of 
all these places, even in those whigh they corre<Sted in their 
Latin translation, have they cancelled the words a Ert/j^e in 
the Greek text, as they ought to have done. 

The same words « srt/x,? arc foundinfour places of Prop. 11 
of this book, in the first and last of which they are necessary, 
but in the second and third, though they are true, they are 
quite superfluous ; as they likewise are in the second of the two 
places in which they are found in the 12th Prop, and in the 
like places of Prop. 22, 23, of this book ; but are wanting in 
the last place of Prop. 23, as also in Prop. 25, Book 11. 

COR. PROP. IV. B. V. 

This Corollary has been unskilfully annexed to this propo- 
sition, afid has been made instead of the legitimate demon- 
stration, which, without doubt, Theon, or some other editor, 
has taken away, not from this, but from its proper place in 

this 



NOTES. 307 

this book: The author of it designed to demonstrate, that if Book v. 
four magnitudes E, G, F, H, be proportionals, they are also ^•^''^^ 
proportionals inversely ; that is, G is to E, as H to F j which 
is true ; but the demonstration of it does not in the least depend 
upon this 4th prop, or its demonstration : For, when he says, 
*' because it is demonstrated, that if K be greater than M, L is 
** greater than N," &c. This indeed is shewn in the demon- 
stration of the 4th prop, but not from this, that E, G, F, H, 
are proportionals ; for this last is the conclusion of the propo- 
sition. Wherefore these words, " because it is demonstrated," 
&c. are wholly foreign to his design : And he should have 
proved, that if K be greater than M, L is greater than N, from 
this, that E, G, F, H, are proportionals, and from the 5th 
def. of this book, which he has not ; but is done in proposition 
B, which we have given in its proper place, instead of this co- 
rollary i and another corollary is placed after the 4th prop, 
which is often of use ; and is necessary to the demonstration 
of Prop. 18 of this book. 



PROP. V. B. V. 

In the construftion which precedes the demonstration of 
this proposition, it is required that EB may be the same mul- 
tiple of CG, that AE is of CF ; that is, that EB be divided 
into as many equal parts, as there are parts in AE equal to 
CF : From which it is evident, that this construction is not 
Euclid's, for he does not shew the way of dividing straight 
lines, and far less other magnitudes, into any number of equal 
parts, until the 9th proposition of B. 6. and he nevpr requires 
any thing to be done in the construction of which he had not 
before giren the method of doing : For this rea- 
son, we have changed the construdlion to one. 



C 



which, without doubt, is Euclids, in which no- 
thing is required but to add a magnitude to itself 
a certain number of times ; and this is to be found 
in the translation from the Arabic, though the 
enunciation of the proposition and the demonstra- 
tion are there very much spoiled. Jacobus Peleta- 
rius, who was the first, as far as I know, who took B^ J) 
notice of this error, gives also the right construc- 
tion in his edition of Euclid, after he had given the other which 
he blames : He says, he would not leave it out, because it was 
fine, and might sharpen one's genius to invent others like it ; 
' X 2 whereas 



3o8 NOTES. 

Book V, whereas thercis not the least difference between the twodemon- 

■^**''"''"*'^ strations, except a single word in the constru^lion, which very 

probably has been owing to an unskilful librarian. Claviuslike- 

wise gives both the ways ; but neither he nor Peletarius takes 

notice of the reason why the one is preferable to the other. 

PROP. VI. B. V. 

There are two cases of this proposition, of which only the 
first and simplest is demonstrated in the Greek : And it is pro- 
bable Theon thought it was sufficient to give this one, since 
he was to make use 'of neither of them in his mutilated edition 
of the 5th book ; and he might as well have lett out the other, 
■ as also the fifth proposition, for the same reason. The demon- 
stration,ot the other case is now added, because both of them, 
as also the 5th proposition, are necessary to the demonstration 
of the i8th proposition of this book. The translation from 
the Arabic gives both cases briefly. 

PROP. A. B. V. 

This proposition is frequently used by geometers, and it is 
necessary in the 25th prop, of this book, 31st of the 6th, and 
34th of the nth, and 15th of the 12th book : It seems to have 
been taken out of the Elements by Theon, because it appeared 
evident enough to him, and others, who substitute the confused 
and indistindl idea the vulgar have of proportionals, in place 
of that accurate idea which is to be got from the 5th def. of 
this book. Nor can there be any doubt that Eudoxus or Euclid 
gave it a place in the Elements, when we see the 7th and 9th 
of the same book demonstrated, though they are quite as easy 
and evident as this. Alphonsus Borellus takes occasion from 
this proposition to censure the 5th definition of this book very 
severely, but most unjustly. In p. 126 of his Euclid Restored, 
printed at Pisa in 1658, he says, " Nor can even this least 
** degree of knowledge be obtained from the foresaid property," 
viz. that which is contained in 5th def. 5. '' That, if four 
*' magnitudes be proportionals, the third must necessarily be 
" greater than the fourth, when the first is greater than the 
*' second : as Clavius acknowledges in the i6th prop, of the 
** 5th book of the elements." But though Clavius makes no 
such acknowledgment expressly, he has given Borellus a han- 
dle to say this of him; because -Cvhen Clavius, in the above 
cited place, censures Commandine, and that very justly, for 
demonstrating this proposition by help of the 16th of the 5th j 
yet he himself gives no demonstration of it, but thinks it plain 

from 



NOTES. 



309 



from the nature of proportionals, as he writes in the end of the Book. v. 
14th and i6:h Prop. B. 5. of his edition, and is followed by ^^**'**', 
Herigoa in Schol. I. Prop. 4. B. 5, as if there was any nature 
of proportionals antecedent to that which is to be derived and 
understood from the definition of them : And indeed, though 
it is very easy to give a right demonstration of it, nobody, as 
far as I know, has given one, except the learned Dr. Barrow, 
who, in answer to Borellus's objection, demonstrates it indi- 
rectly, but very briefly and clearly, from the 5th definition, 
in the 322d page of his Lect. Mathem, from which definition 
it may also be easily demonstrated directly : On which ac* 
count we have placed it next to the propositions concerning 
equimultiples. 

PROP. B. B. V. 

This also is easily deduced from the 5th def. B. 5, and 
therefore is placed next to the other ; for it was very igno- 
rantly made a corollary from the fourth Prop, of this Book. 
See the note on that corollary. 

PROP. C. B. V. 

This is frequently made use of by geometers, and is neces- 
sary to the 5th and 6th Propositions of the loth Book. Cla- 
vius, in his notes subjoined to the 8tli def. of Book 5. demon- 
strates it only in numbers, by help of some of the propositions 
of the 7th Book: in order to demonstrate the property con- 
tained in the 5th definition of the 5ch Book, when applied to 
numbers, from the property of proportionals contained in the 
20th def. of the 7th Book : And most of the commentators 
judge it difficult to prove that four magnitudes which are pro- 
portionals according to the 20th def. of 7th Book, are also pro- 
portionals according to the 5th def. of 5th Book. But this 
is easily made out as follows : 

First, if A, B, C, D, be four mag- 
nitudes, such that A is the same mul- 
tiple, or the same part of B, vihich 
C is of D : A, B, C, D, are propor- 
tionals : This is demonstrated in pro- 
position C. 

Secondly, if AB contain the same 
parts of CD that EF does of GH j in 
this case likewise AB is to CD, as EF 
to GH. 

X3 



B 



D 



R 



ACE 



H 

L 

G 

Let 



310 , NOTES. 

Book V. Let CK be a part of CD, and GL the same part of GH ; 

^"**v*^ and let AB be the same multiple of 

CK, -:hat EF is of GL : Therefore, F 

by Prop. C, of 5th Book, AB is to B TT 

CK, as EF to GL : And CD, GH jy 

are equimultiples^ of CK, GL the se- 
cond and fourth j wherefore, by Cor. 
Prop. 4. Book 5. AB is to CD, as EF K 

to GH. 

And if four magnitudes be propor- A P E Gl 
tionals according to the 5th def. of 
Book 5, they are also proportionals according to the 20th def. 
of Book 7. 
' First, if A be to B, as C to D ; then if A be any multiple 
or part of B, C is the same multiple or part of D, by Prop. D. 
of B. 5. 

Next, if AB be to CD, as EF to GHj then if AB con- 
tainsany partsofCD, EF contains the same parts of GH: 
For let CK be a part of CD, and GL the same part of 
, GH, and let AB be a multiple of CK : EF is the same 

multiple of GL : take M the same multiple of GL that 
AB is of CK ; therefore, by Prop. C. of B. 5. AB is to 
CK, as M toGL : and CD, GH are equimultiples of 
CK, GL ; wherefore, by Cor. Prop. 4. B. 5. AB is to 
CD, as M to GH. And, by the hypothesis, AB is to 
CD, as EF to GH ; therefore M is equal to EF by M 
Prop. 9. Book 5. and consequently EF is the same multiple of 
GLthatABisofCK. 

PROP. D. B. V. 

This is not unfrequently used in the demonstration of other 
' propositions, and is necessary in that of Prop. 9. B. 6. It seems 
Theon has left it out for the reasons mentioned in the notes 
at Prop. A. 

PROP. Viir. B. V. 

Li the demonstration of this, as it is now in the Greek, 
there are two cases (see the demonstration in Hervagius, or 
Dr. Gregory's edition), of which the first is that in which AE 
is less than EB ; and in this it necessarily follows, that H© 
the multiple EB is greater than ZH the same multiple of AE, 
which last multiple, by the construction, is gteater than A j 
whence also H0 must be greater than A : But in the second 
case, viz. that in which EB is less than AE, though ZH be 
greater than A, yet n© may be less than the same A ; so that 

there 



NOTES. 



3" 



H 



A 

E- 



HF 



e B A © 



there cannot be taken a multiple of a which is the first that is ^°°^ v* 
greater than K or H0, because A itself is greater than it : ^"^^'^^ 
Upon this account, the author of this demonstration found it 
necessary to change one part of the construction that was made 
use of in the first case : But he has, without any necessity, 
changed also another part of it, viz. when he orders to tak« 
N that multiple of a which 
is the first that is greater than 
ZH } for he might have taken 
that multiple of a which is 
the first that is greater than 
H©, or K, as was done in the 
first case: Helikewisebrings 
in this K into the demonstra- 
tion of both cases, without 
any reason; for it serves to no 
purpose but to lengthen the 
demonstration. There is also 

a third case which is not mentioned in this demonstration, viz. 
that in which AE in the first, or EB in the second of the two 
other cases, is greater than D ; and in this any equimultiples, 
as the doubles, of AE, EB are to be taken, as is done in this 
edition, where all the cases are it once demonstrated : And 
from this it is plain that Theon, or some other unskilful editor, 
has vitiated this proposition. 

PROP. IX. B. V. 

Of this there is given a more explicit demonstration than 
that which is now in the Elements. 

PROP.X. B. V. 

It was necessary to give another demonstration of this pro- 
position, because that which is in the Greek and Latin, or 
other edition?, is not legitimate : For the words greater, the 
same, or equal, lesser, have a quite different meaning when ap- 
plied to magnitudes and ratios, as is plain from the 5th and 
7th definitions of Book 5. By the help of these let us exa- 
mine the demonstration ot the loth Prop, which proceeds thus : 
*' Let A have to C a greater ratio than B to C : I say that A is 
" greater than B ; for if it is not greater, it is either equal or 
" less. But A cannot be equal to B, because then each of them 
*' would have the same ratio to C; but they have not. There- 
" fore A is not equal to B." The force of whichreasoning is 
this : If A had to C the same ratio that B has to C, then if 

X 4 any 



3r2 NOTES. 

Book V. any equimultiples whatever of A and B be taken, and any 
^"^'^^^^ multiple whatever of C j if the multiple of A be greater than 
the multiple of C, then, by the 5th def. of Book 5, the multiple 
of B is also greater than that of C : but, from the hypothesis 
that A has a greater ratio to C, than B has to C, there must, 
by the 7th def. of Book 5, be certain equimultiples of A and B, 
and some multiple of C such, that the multiple of A is greater 
than the multiple of C, but the muhiple of B is not greater 
than the same multiple of C : And this proposition direflly 
contradi6ls the preceding ; wherefore A is not equal to B. 
The demonstration of the loth pcop. goes on thus: *' But 
" neither is A less than B ; because then A would have a less 
" ratio to C than B has to it : But it has not a less ratio, there- 
*' fore A is not less than B," &c. Here ijt is said, that " A 
** would have a less ratio to C than B has to C," or, which is 
the same thing, that B would have a greater ratio to C than 
A to C ; that is, by 7 th def. Book 5, there must be some equi- 
multiples ef B and A, and some multiple of C, such that the 
multiple of B is greater than the multiple of C, but the mul- 
tiple of A is not greater than it : And it ought to have been 
proved, that this can never happen if the ratio of A to C be 
greater than the ratio of B to C ; that is, it should have been 
proved, that, in this case, the multiple of A is always greater 
than the multiple of C, whenever the multiple of B is greater 
than the multiple of C j for when this is demonstrated, it will 
be evident that B cannot have a greater ratio to C, than A has 
to C, or, which is the same thing, that A cannot have a less 
ratio to C than B has to C . But this is not at all proved in the 
loth proposition : But if the lOth were once demonstrated, it 
would immediately follow from it, but cannot without it be 
easily demonstrated, as he that tries to do it will find. Where- 
fore the lOth proposition is not sufficiently demonstrated. And 
it seems that he who has given the demonstration of the loth 
proposition as we now have it, instead of that which Eudoxus 
or Euclid had given, has been deceived in applying what is 

, manifest, when understood of magnitudes, unto ratios, viz. that 

a magnitude cannot be both greater and less than another. 
That those things which are equal to the same are equal ta 
one another, is a most evident axiom when understood of 
magnitudes; yet Euclid does not make use of it to infer, that 
those ratios which are the same to the same ratio, are the same 
to one another; but explicitly demonstrates this in Prop. ii. of 
Book 5. The demonstration we have given of the i oth prop. 

is 



NOTES. 



313 



If A have to C a 



A 
D 



E 



is no doubt the same with that of Eudoxus or Euclid, as it is ^°^^ 
immediately and directly derived from the definition, of a ^^ 
greater ratio, viz. the 7 of the 5. 

The above-mentioned proposition, viz. 
greater ratio than B to C j and if of A 
and B there be taicen certain equimulti- 
ples, and some multiple of C ; then if the 
multiple of B be greater than the multiple 
of C, the multiple of A is also greater 
than the same, is thus demonstrated : 

LetD, E be equimultiples of A, B, and 
F a multiple of C, such, that E the mul- 
tiple of B is greater than F j D the mul- 
tiple of A is also greater than F. 

Because A has a greater ratio to C , than 
B to C, A is greater than B, by the loth 
Prop. B. 5, therefore D the multiple of A 
is greater than E the same mulciple of 
B : And E is greater than F : much 
more therefore D is greater than F. 



c 



PROP. XIII. B. V. 

In Commandine's, Briggs's, and Gregory's translations, at 
the beginning of this demonstration, it is said, " And the mul- 
'' tiple of C is greater than the multiple of Dj but the mul- 
" tiple of E is not greater than the multiple of F :" Which 
words are a literal translation from the Greek : But the sense 
evidently requires that it be read, " so that the multiple of C 
*' be greater than the multiple of D ; but the multiple of E be 
"not greater than the multiple of F." And thus this place 
was restored to the true reading in the first editions of Com- 
mandine's Euclid, printed in 8vo. at Oxford : But in the later 
editions, at least in that of 1747, the error of the Greek text 
was kept in. 

There is a corollary added to prop. 13, as It Is necessary to 
the 20th and 21st prop, of this book, and is as useful as the 
proposition. 



PROP. XIV. B. V. 

The two cases of this, ^hlch are not in the Greek, are 
added ; the demonstration of them not being exactly the same 
with that of the first case. 




NOTES. 

PROP. XVII. B. V. 

The order of the words in a clause of this Is changed to 
one more natural : As was also done in pfop, ii. 

PROP. XVIII. B. V. 

The demonstration of this is none of Euclid's, nor is it legi- 
timate ; for it depends upon this hypothesis, that to any three 
magnitudes, two of which,- at least, are of the same kind, 
there may be a fourth proportional : which, if not proved, the 
demonstration now in the text is of no force : But this is 
assumed without any proof; nor can it, as far as I am able to 
discern, be demonstrated by the propositions preceding this : 
so far is it from deserving to be reckoned an axiom, as Cla- 
vius, after other commentators, would have it, at the end of 
the definitions of the 5Ch book. Euclid does not demonstrate 
it, nor does he shew how to find the fourth proportional, 
before the 12th prop, of the 6th book : And he never assumes 
any thing in the demonstration of a proposition, which he had 
not before demonstrated ; at least, he assumes nothing the exist- 
ence of which is not evidently possible; for a certain conclusion 
can never be deduced by the means of an uncertain proposition : 
Upon this account, we have given a legitimate demonstration 
of this proposition instead of that in the Greek and other edi- 
tions, which very probably Theon, at least some other, has 
put in the place of Euclid's, because he thought it too prolix: 
And as the 17th prop, of which this i8th is the converse, is 
demonstrated by help of the 1st and 2nd propositions of this 
book ; so, in the demonstration now given of the 1 8th, the 5th 
prop, and both cases of the 6th are necessary, and these two 
propositions are the converses of the ist and 2d. Now the 5th 
arid 6th do not enter into the demonstrationof any proposition 
in this book as we now have it : Nor can they be of use in any 
proposition of the Elements, except in this i8th, and this is a 
manifest proof, that Euclid made use of them in his demon- 
stration of it, and that the demonstration now given, which is 
exactly the converse of that of the 17th, as it ought to be, dif- 
fers nothing from that of Eudoxus or Euclid : For the 5th and 
6th have undoubtedly been put into the 5th book for the sake 
of some propositions in it, as all the other propositions about 
equimultiples have been. 

Hieronymus Saccherius, in his book named " Euclides ab 
"omni nasvo vindicatus," printed at Milan ann. 1733, in 410, 

acknowledge s 



NOTES. 315 

acknowledges this blemish in the demonstration of the i8th, ^^^^^^ 
and that he may remove it, and render the demonstration we 
now have of it legitimate, he endeavours to denwnstrate the 
following proposition, which i-s in page 115 of his book, viz. 

*' Let A, B, C, D be four magnitudes, of which the two 
" first are of the one kind, and also the two others either of the 
" same kind with the two first, or of some other the same 
*' kind with one another. I say the ratio of the third C to the 
" fourth D, is either equal to, or. greater, or less than the ratio 
"of the first A to the second B." 

And after two propositions premised as lemmas, he proceeds 
thus ; 

** Either among all the possible equimultiples of the first 
A, and of the third C, and, at the same time, among all 
the possible equimultiples of the second B, and of the 
fourth D, there can be found some one multiple EF of the 
first A, and one IK of the second B, that are equal to one 
another j and also (in the s.cme casQ) some one multiple 
GH of the third C equal to LM the multiple of the fourth 
D, or such equality is no where to be found. If the first 
case happen, 

[i. e. if such A £ f 

equality is to 

be found] it B 1 ^K_ 

is manifest 

from what is C G H 

before demon- 
strated, that J) J^ J£ 

A is to B 

as C to D } but if such simultaneous equality be not to be 
found upon both sides, it will be found either upon one 
side, as upon the side of A [and B] ; or it will be found 
upon neither side j if the first happen : therefore (from 
Euclid's definition of greater and lesser ratio foregoing) 
A has to B a greater or less ratio than C to D ; according 
as GH the multiple of the third C is less, or greater 
than LM the multiple of the fourth D : But if the second 
case happen ; therefore upon the one side, as upon the side 
of A the first and B the sec6nd, it may happen that the 
multiple EF [viz. of the first] may be less than IK the 

* multiple of the second, while, on the contrary, upon the, 

* other side [viz. of C and D], the multiple GH [of the third 
'C]is greater than the other multiple LM [of the fourth 

** D] : And then (from the same definition of Euclid) the ratio 

"of 



3i6 NOTES. 

Book V' "of the first A to the second B, is less than the ratio of the 
^"^""'f^^ <« third C to the fourth D j or on the contrary. 

*' Therefore the axiom [i. e. the proposition before set down] 
"remains demonstrated," &c. 

Not in the least ; but it still remains undemonstrated j For 
what he says may happen, may, in innumerable cases, never 
happen; and therefore his demonstration does not hold: For 
example, if A be the side, and B the diameter of a square ; 
and C the side, and D the diameter of another square ; there 
can in no case be any multiple of A equal to any of B ; nor 
any one of C equal to one of D, as is well known ; and yet 
it can never happen that when any multiple of A is greater 
- than a multiple of B, the multiple of C can be less than the 
multiple of D, nor when the multiple of A is less than that of 
B, the multiple of C can be greater than that of D, viz. talcing 
equimultiples of A and C, and equimultiples of B and D : For 
A, B, C, D are proportionals j, and so if the multiple of A be 
greater, &c. than that of B, so must that of C be greater, &c. 
than that of D ; by 5th Def. b. 5. 

The same objection holds good against ?he demonstration 
which some give of the first prop, of the 6th book, which we 
have made against this of the i8th prop, because it depends 
upon the same insufficient foundation with the other, 

PROP. XIX. B. V. 

A COROLLARY is added to this, which is as frequently used 
as the proposition itself. The corollary which is subjoined to it 
in the Greek, plainly shews that the 5th book has been vitiated 
by editors who were not geometers : For the conversion of 
ratios does not depend upon this igyth, and the demonstration 
which several of the commentators on Euclid give of conver- 
"* sion is not legitimate, as Clavius has rightly observed, who 
has given a good demonstration of it which we have put i^ 
proposition E ; but he makes it a corollary from the 19th, and 
begins it with the words, " Hence it easily follows," though 
it does not at all follow from it. 

PROP. XX. XXL XXII. XXIII. XXIV. B. V. 

The demonstration of the 20th and 21st propositions, are 
shorter than those Euclid gives of easier propositions, either 
in the preceding or following books : Wherefore it was pro- 
per CO make them more explicit, and the 22d and 23d propo- 
sLons arc, as they ought to bcj extended to any number of 

magnitudes : 



NOTES. 317 

magnitudes : And, in like manner, may the 24th be, as is Boos V. 
taken notice of in the corollary j and another corollary is added, ^"^"^"'^'^ 
as useful as the proposition, and the words, " any whatever" 
are supplied near the end of prop. 23, which are wanting in 
the Greek text, and the translations from it. 

In a paper writ by Philippus Naudaeus, and published after 
his death, in the history of the Royal Academy of Sciences of 
Berlin, anno 1745, page 50, the 23d prop, of the 5th book is 
censured as being obscurely enunciated, and, because of this, 
prolixly demonstrated : The enunciation there given is not 
Euclid's, but Tacquet's, as he acknowledges, which, though 
not so well expressed, is, upon the matter, the same with that 
which is now in the Elements. Nor is there any thir\g ob- 
scure in it, though the author of the paper has set down the 
proportionals in a disadvantageous order, by which it appears 
to be obscure : But no doubt Euclid enunciated this 23d, as 
well as the 22d, so as to extend it to any number of magni- 
tudes, which, taken two and two, are proportionals, and not 
of six only ; and to this general case, the enunciation which 
Naudaeus gives, cannot be well applied. 

The demonstration which is given of this 23d, in that paper, 
is quite wrong ; because, if the proportional magnitudes be 
plane or solid figures, there can no rectangle, (which he impro- 
perly calls a produii) be conceived to be made by any two of 
them : And if it should be said, that in this case straight lines 
are to be taken which are proportional to the figures, the de- 
monstration would this way become much longer than Eu- 
clid's : But, even though his demonstration bad been right, who 
does not see that it could not be made use of in the 5th book i 



PROP. F, G, H,K. B. V. 

These propositions are annexed to the 5th book, because 
they are frequently made use of by both ancient and modern 
geometers : And in many cases, compound ratios cannot be 
brought into demonstration, without making use of them. 

Whoever desires to see the dodlrine of ratios delivered in 
this 5th book solidly defended, and the arguments brouirht 
against it by And. Tacquet, Alph. Borellus and others, fully 
refuted, may read Dr. Barrow's mathematical lectures, viz. the 
7th and 8th of the year 1666. 

The 5th book being thus corrected, I most readily agree to 

Wh^t the learned Dr. Barrow says,* " That there is nothing 

I «in 

♦Page 336. 



3i8 NOTES; 

" in the whole body of the Elements of a more subtile inven- 
" tion, nothing more solidly established, and more accurately 
*' handled, than the do£lrine of proportionals." And there is 
some ground to hope, that geometers will think that this could 
not have been said with as good reason,- since Theon's time 
till the present. ^ ^ 



Book VI. 



DEF. II. and V. ofB. VI. 

X HE 2d definition does not seem to be Euclid's but some 
unskilful editor's : For there is no mention made by Euclid, 
nor, as far as I know, by any other geometer, of reciprocal 
figures: It is obscursly expressed, which made it proper to 
render it more distinct : It would be better to put the follow- 
ing definition in place of it, viz. 

DEF. II. 

Two magnitudes are said to be reciprocally proportional to 
two others, when one of the first is to one of the other magni- 
tudes, as the remaining one of the last two is to the remaining 
one of the first. 

But the 5th definition, whicb, since Theon's time has been 
kept in the elements, to the great detriment of learners, is 
now justly thrown out of them, for the reasons given in the 
notes on the 23d prop, of this book. 

PROP. I. and II. B. VI. 

To the first of these a corollary is added, which is often 
Ksed : And the enunciation of the second is made more 
general. 

PROP. Ill B. VI. 

A SECOND case of this, as useful as the first, is given in 
prop. A; viz. the case in which the exterior angle of a trian- 
gle is bise<5led by a straight line : The demonstration of it is . 
very like to that of the first case, and upon this account may, 
probably, have been left out, as also the enunciation, by some 
unskilful editor. At least, it is certain, that Pappus makes 
use of this case, as an elementary proposition, without a de- 
monstration of it, in Prop. 39, of his 7th Book of Mathema- 
tical Colledtions. 



NOTES. 

PROP. VII. B. VI. 

To this a case is added which occurs not unfrcquentljr in 
demonstration. 

PROP. VIII. B. VI. ' 

It seems plain that some editor has changed the demonstra- 
tion that Euclid gave of this proposition : For, after he has 
demonstrated, that the triangles are equiangular to one ano- 
ther, he particularly shews that their sides about the equal 
angles are proportionals, as if this had not been done in the 
demonstration of the ^th prop, of this book : This superfluous 
part is not found in the translation from the Arabk, and is 
now left out. 

PROP. IX. B. VI. 

This is demonstrated in a particular case, viz. that in which 
the third part of a straight line is required to be cut ofF;. which 
is not at all like Euclid's manner : Besides, the author of thi 
demonstration, from four magnitudes being proportionals, con 
eludes that the third, of them is the same multiple of the fourth, 
which the first is of the second j now, this is no where demon- 
strated in the 5th book, as we now have it : But the editor 
assumes it from the confused notion which the vulgar have of 
proportionals : On this account it was necessary to give 
general and legitimate demonstration of this proposition. 




e 



a 



PROP. XVIII. B. VI. 

The demonstration of this seems to be vitiated : For the 
proposition is demonstrated only in the case of quadrilateral 
iigures, without mentioning how it may be extended to figures 
ot five or more sides : Besides, from two triangles being equi- 
angular, it is inferred, that a side of the one is to the homolo- 
gous side of the other, as another side of the first is to the side 
homologous to it of the other, without permutation of the 
proportionals ; which is contrary to Euclid's manner, as is 
clear from the next proposition: And the same fault occurs 
again in the conclusion, where the sides about the equal angles 
are not shewn to be proportionals, by reason of again neglefl- 
ing permutation. On these accounts, a demonstration is given 
in Euclid's manner, like to that he makes use of in the 20th 
3 prop. 



-320 NOTES. 

Book VI. prop, of this boolc ; and it is extended to five- sided figures, by 
'^^''^^'^^ which it may be seen how to extend it to figures of any num- 
ber of sides. 

PROP. XXIII. B. VI. 

Nothing is usually reckoned more difficult in the elements 
of geometry by learners, than the doctrine of compound ratio, 
which Theon has rendered absurd and ungeometrical, by sub- 
stituting the 5th definition of the 6th book in place of the 
right definition, which without doubt Eudoxus or Euclid gave, 
in its proper place, after the definition of triplicate ratio, 
&c. in the 5th book. Theon's definition is this ; a ratio is 
said to be compounded of ratios 'orx* ai TuvyoXuii myiyiMryirts zp* 
sxvras iroXKx'nKxaiaty^itaxt wo/^yfT; t;»« :, Which Commandine thus 
translates : " Quando rationum quantitates inter se multi- 
" plicatae aliquam efficient rationem ;" that is, when the 
quantities of the ratios being multiplied by one another make 
a certain ratio. Dr. Wallis translates the word w^jX/xornTfj 
** rationem exponentes," the exponents of the ratios : And Dr. 
Gregory renders the last words of the definition by " illius 
** facit quantitatem," makes the quantity of that r,atio : But in 
whatever sense tne " quantities," or " exponents of the ratios,'* 
and their '' multiplication," be taken, the definition will be 
ungeometrical and useless : For there can be no multiplication 
but by a number : Now the quantity or exponent of a ratio 
(according to Eutochius in his Comment, on Prop. 4. Book 2, 
of Arch, de Sph. et Cyl. and the moderns explain that term) 
is the number which multiplied into the consequent term of a 
ratio produces the antecedent, or, which is the same thing, the 
■number which arises by dividing the antecedent by the conse- 
quent ; but there are many ratios such, that no number can 
arise from the division of the antecedent by the consequent ; 
ex. gr. the ratio of which the diameter of a square has to the 
side of it } and the ratio which the circumference of a circle has 
to its diameter, and such like. Besides, that there is not the 
least mention made of this definition in the writings of Euclid, 
Archimedes, Apollonius, or other ancients, though they fre- 

. quentJy make use of compound ratio : And in this 23d prop. 

' of the 6th book, where compound ratio is first mentioned, there 
is not one word which can relate to this definition, though 
here, if in any place, it was necessary to be brought in ; but 
the right definition is expressly cited in these words: " But the 
" ratio of K to M is compounded of the ratio of K to L, 

^ " aiid 



NOTES. 321 

" and of the ratio of L to M." This definition therefore of Book vi. 
Theon is quite useless and absurd : For that Theon brought it ""'^■'^^ 
into the Elements can scarce be doubted ; as it is to be found 
in his commentary upon Ptolemy's Mf7«x»> ivirx^is, page 62. 
where he also gives a childish explication of it, as agreeing 
only to such ratios as can be expressed by numbers ; and from 
this place the definition and explication have been exactly co- 
pied and prefixed to the definitions of the 6th book, as appears 
from Hervagius's edition : But Zambertusand Commandine, 
in their Latin translations, subjoin the same to these defini- 
tions. Neither Campanus, nor, as it seems, the Arabic ma- 
nuscripts, from which he made, his translation, have this de- 
finition. Claviiis, in his observations upon it, rightly judges, 
that the definition of compound ratios might have been made 
after the same manner in which the definitions of duplicate 
and triplicate ratio are given, viz. " That as in several magni- 
" tudes that are continual proportionals, Euclid named the 
" ratio of the first to the third, the duplicate ratio of the first 
" to the second ; and the ratio of the first to the fourth, the 
*' triplicate ratio of the first to the second, that is, the ratio 
*' compounded of two or three interjnediate* ratios that are 
" equal to one another, and so on ; so, in like manner, if 
*' there be several magnitudes of the same kind, following one 
" another, which are not continual proportionals, the first is 
*' said to have to the last the ratio compounded of all the inter- 

" mediate ratios, only for this reason, that these inter- 

" mediate ratios are interposed betwixt the two extremes, viz. 
" the first and last magnitudes ; even as, in the loth definition 
" of the 5th book, the ratio of the first to the third was called 
" the duplicate ratio, merely upon account of two ratios 
" being interposed betwixt the extremes, that are equal to one 
*' another : so that there is no difference betwixt this com- 
** pounding of ratios, and the duplication or triplication of 
** them which are defined in the 5th book, but that in the du- 
** plication, triplication, &£c. of ratios, all the interposed ratios 
" are equal to one another j whereas, in the compounding of 
" ratios, it is not necessary that the intermediate ratios should 
*' be equal to one another," Also Mr. Edmund Scarburgh, 
in his English translation of the first six books, page 238, 
266, expressly affirms, that the 5:h definition of the 6th book 
is suppositious, and that the true definition of compound rr.tio 
is contained in the lOth definition of the 5th book, viz. the 

Y deftnition 



322 NOTES. 

BooklVI. definition of duplicate ratio, or to be understood from it, to 
'""^^^"^ wit, in the same manner as Clavius has explained it in the 
preceding citation. Yet these, and the rest of the moderns, 
do notwithstanding retain this 5th def. of the 6:h book, and il- 
lustrate and explain it by long commentaries, when they ought 
rather to have taken it quite away from the Elements. 

For, by comparing def. 5, book 6, with prop. 5, book 8, it 
will clearly appear that this definition has been put into the 
Elements in place of the right one, which has been taken out 
of them : Because, in prop. 5, book 8, it is demonstrated, that 
the plane number of which the sides are C, D, has to the plane 
number of which the sides are E, Z (see Hervagius's or Gre- 
. gory's edition), the ratio which is compounded of the ratios of 
their stdes j that is, of the ratios of C to E, and D to Z ; and 
by de;. 5. book 6, and the explication given of it by all the 
commentators, the ratio which is compounded of the ratios 
of C to E, and D to Z, is the ratio of the product made by 
the multiplication of the antecedents C, D, to the product by 
the consequents E, Z, that is, the ratio of the plane number 
of which the sides are C, D, to the plane number of which the 
sides are E, Z. Wherefore the proposition which is the 5th 
def. of book 6, is the very same with the 5thprop. of book 8, 
and therefore it ought necessarily to be cancelled in one of 
these places ; because it is absurd that the same proposition 
should stand as a definition in one place of the Elements, and 
be demonstrated in another place of them. Now, there is no 
doubt that prop. 5, book 8, should have a place in the Elements, 
as the same thing is demonstrated in it concerning plane num- 
bers, which is demonstrated in prop. 23, book 6, of equiangu- 
lar parallelograms ; wherefore def. 5, book 6, ought not to b? 
in the Elements. And from this it is evident that this defi- 
nition is not Euclid's, but Theon's, or some other unskilful 
geometer's. 

But nobody, as far as I know, has hitherto shewB the true 
use of compound ratio, or for what purpose it has been intro^ 
duced into geometry; for every proposition in which com- 
pound ratio is made use of, may without it be both enun- 
ciated and derr\onstrated. Now the use of compound ratio 
consists wholly in this, that by means of it, circumlocutions 
piay be avoided, and thereby propositions may be more briefly 
either enunciated or demonstrated, or both may be done; for 
instance, if thi? 23d proposition of the sixth book were to be 
enunciated, without mentioning compound ratio, it might be 

done 



NOTES. 323 

done as follows : If two parallelograms be equiangular, and ^'^^ ^^• 
if as a side of the first to a side of the second, so any assumed ^■*''"^'"*^ 
straight line be njade to a second straight line j and as the 
other side of the first to the other side of the second, so the 
second straight line be made to a third. The first parallelo- 
gram is to the second, as the first straight line to the third. 
And the demonstration would be exactly the same as we now 
have it. But the ancient geometers, when they observed this 
enunciation could be made shorter, by giving a name to the 
ratio which the first straight line has to the last, by which 
name the intermediate ratios might likewise be signified, of 
the first to the second, and of the second to the third, and so 
on, if there were more of them, they called this ratios of the 
first to the last, the ratio compounded of the ratio of the first 
to the second, and of the second to the third straight line; 
that is, in the present example, of the ratios which are the 
same with the ratios of the sides, and by this they expressed 
the proposition more briefly thus : If there be two equiangular 
parallelograms, they have to one another the ratio which is the 
same with that which is compounded of ratios that are the 
same with the ratios of the sides ; which is shorter than the 
preceding enunciation, but has precisely the same meaning. 
Or yet shorter thus : Equiangular parallelograms have to one 
another the ratio which is the same with that which is com- 
pounded of the ratios of their sides. And these two enuncia- 
tions, the first especially, agree to the demonstration which is 
now in the Greek. The proposition may be more briefly de- 
monstrated, as Candalla does, thus : Let ABCD, CEFG be two 
equiangular parallelograms, and complete the parallelogram 
CDHG i then, because there are three parallelograms, AC, 
CH, CF, the first AC (by the definition of compound ratio) 
has to the third CF, the ratio which 
is compounded of the ratio of the first 
AC to the second CH, and of the 
ratio of CH to the third CF ; but 
the parallelogram AC is to the pa- 
rallelogram CH, as the straight line 
BC to CG ; and the parallelo- 
gram CH is to CF, as the straight 

line CD is to CE j therefore the parallelogram AG has to CF 
the ratio which is compounded of ratios that are the same with 
the ratios of the sides. And to this demonstration agrees the 
enunciation which is at present in the text, viz. equiangular 
Y z parallelograms 




324 N O f E S. 

Book VI. parallelograms have tooheanothertheratio which is compound- 
^'^^^''^ ed of the ratios of the sides : for the vulgar reading, " which 
** is compounded of their sides," is absurd. But, in this edi- 
tion, we have kept the demonstration which is in the Greek 
text, though not so short as Candalla's ; because the way of 
finding the ratio which is compounded of the ratio of the sides, 
that is, of finding the ratio of the parallelograms, is shewn in 
that, but not in Candalla's demonstration ; whereby beginners 
may learn, in like cases, how to find the ratio which is com- 
pounded of two or mc;re given ratios. 

P'rom what has been said, it may be observed, that in any 
magnitudes whatever of the same kind A, B, C, D, &c. the 
ratio compounded of the ratios of the first to the second, of 
the second to the third, and soon to the last, is only a name or 
expression by which the ratio which the first A has to the last 
D is signified, and by which at the same time the ratios of all 
the magnitudes A to B, B to C, Cto D, from the first to the 
last, to one another, whether they be the same, or be not the 
same, are indicated ; as in magnitudes which are continual 
proportionals A, B, C, D, &c. the duplicate ratio of the first 
to the second is only a name, or expression by which the ratio 
of the first A to the third C is signified, and by which, at the 
same time, is shewn, that there are two ratios of the magni- 
tudes from the first to the last, viz. of the first A to the second 
B, and of the second B to the third or last C, which are the 
same with one another ; and the triplicate ratio of the first to 
the second is a name or expression by which the ratio of the 
first A to the fourth D is signified, and by which, at the same 
time, is shewn, that there are three ratios of the magnitudes 
from the first to the last, viz. of the first A to the second B, 
and of B to the third C, and of C to the fourth or last D, 
which are all the same with one another; and so in the case 
of any other multiplicate ratios. And that this is the right 
explication of the meaning of these ratios is plain from the defi- 
nitions of duplicate and triplicate ratio, in which Euclid makes 
use of the word Xtycritiy is said to be, or is called ; which word 
he, no doubt, made use of also in the definition of compound 
ratio, which Theon, or some other, has expunged from the 
Elements; for the very same word is still retained in the 
wrong definition of compound ratio, which is now the 5th of 
the 6th book : But in the citation of these definitions it is 
sometimes retained, as in the demonstration of prop. 19, book 

6, 



NOTES. 325 

6, **'the first is said to have, "fxiiy keyirxij to the third the du- Book. vi. 

** plicate ratio," &c. which is wrong translated by Comman- ^<^^^^-' 

dine and otherSj *' has," instead of " is said to have :" and 

sometimes it is left out, as in the demonstracion of prop. 33, of 

the iith book, in which we find, "the first has, l^uy to the 

" third the triplicate ratio;" but without doubt l^u., "has," 

in this place, signifies t'ps same as ij^t^, ^eysrxiy is said to have ; 

so likewise in Prop. 23, B. 6, we iiud this citation, " but the 

*' ratio of K to M is compounded, c-vyxnTxi of the ratio of K to ' 

** L, and the ratio L to M," which is a shorter way of 

expressing th^^ same thing, which, according to the definition, 

ought to have been expressed by ovyx«/o-3»< ^eytrxij is said to be 

compounded. 

From these remarks, together with the propositions sub- 
joined to the 5th book, all that is found concerning compound 
ratio, either in the ancient or modern geometers, may be un- 
derstood and explained. 

PROP. XXIV. B. VI. 

It seems that some unskilfiil editor has made up this demon 
stration as we now have it, out ef two others ; one of which- 
may be made from the 2d prop, and the other from the 4th of 
this book. For after he has, from the 2d of this book, and 
composition and permutation, demonstrated, that th jsidesabout 
the angle common to the two parallelograms are proportionals, 
he might have immediately concluded, that the sides about the 
otherequal anoles were proportionals, viz. from Prop. 34, B. i, 
and Prop. 7, B. 5. This he does not, but proceeds to shew, 
that the triangles and parallelograms are equiangular : and in a 
tedious way, by help of Prep. 4. of this book, and the 22d of 
book 5, deduces the same conclusion: FrOm which it is plain, 
that this ill-composed demonstration is not Euclid's : These 
superfluous things are now left out, and a more simple demon- 
stration is given from the 4ih prop, of this book, the same 
which is in the translation from the Arabic, by help of the 2d 
prop, and composition ; but in this the author neglects permu- 
tation, and does not shew the parallelograms to be equiangular, 
as is proper to do for the sake of beginners. 

PROP. XXV. B. VI. 

It is very evident that the demonsiiation which Euclid had 

given of this proposition has been vitiated by some unskilful 

hand : For, after this editor had demonstrated, that, " as the 

•* rectilineal figure ABC is to the rectilineal KGH, so is the 

Y 3 " parallelogram 



3i6 NOTES. 

Book VI. " parallelogram BE to the parallelogram EF ;" nothing more 
^*^v*-' should have been added but this, ^* and the rectilineal figure 
' M ABC is equal to the parallelogram BE ; therefore the recti- 

** lineal KGH is equal to the parallelogram EF," viz. from 
prop. 14, book 5. But betw^ixt these two sentences he has 
inserted this ; " wherefore, by permutation, as the rectilineal 
" figurcABCtothe parallelogram BE, so is the rectilineal KGH 
** to the parallelogram EF ; by which, it is plain, he thought 
it was not so evident to conclude, that the second of four pro- 
portions is equal to the fourth from the equality of the first 
and third, which is a thing demonstrated in the 14th prop, of 
B. 5, as to conclude that the third is equal to the fourth, from 
the equality of the first and second, which is no where demon- 
strated in the Elements as we now have them : But though this 
proposition, viz. the third of four proportionals is equal to the 
fourth, if the first be equal to the second, had been given in 
the Elements by Euclid, as very probably it was, yet he would 
not have made use of it in this place ; because, as was said, the 
conclusion could have been immediately deduced without this 
superfluous step by permutation : This we have shewn at the 
greater length, both because it affords a certain proof of the 
vitiation of the text of Euclid ; for the very same blunder is 
found twice in the Greek text of prop. 23, book ii, and twice 
in prop. 2, book 12, and in the 5, 1 1, 12, and 18th of that book ; 
in which places ©f book 12, except the last of them, it is rightly 
left out in the Oxford edition of Command ine's translation ; 
And also that geometers may beware of making use of permu- 
tation in the like cases : for the moderns not unfrequently com- 
mit this mistake, and among others Commandine himself in his 
commentary on prop. 5, book 3, p. 6, b. of Pappus Alexandri- 
nus, and in other places : The vulgar notion of proportionals 
has, it seems, pre-occupied many so much, that they do not 
sufficiently understand the true nature of them. 

Besides, though the rectilineal figure ABC, to which ano- 
ther is to be made similar, may be of any kind whatever; yet 
in the demonstration the Greek text has "triangle" instead of 
*' reftilineal figure," which error is corrected in the above- 
named Oxford edition. 

PROP. XXVII. B. VI. 

The second case of this has axx^r, otherwise, prefixed ta 
it, as if it was a different demonstration, which probably has 
been done by some unskilful librarian. Dr. Gregory has 

rightly 



NOTES. 



327 



rightly left it oat : The scheme of this second case ought to ^^^ ^'i- 
be marked with the same letters of the alphabet which arc in '*"*^ 
the scheme of the first, as is now done. 



PROP. XXVIII, and XXIX. B. VI. 

These two problems, to the first of which the 27th prop, 
is necessary, are the most general and useful of all in the Ele- 
ments, and are most frequently made use of by the ancient 
geometers in the solution of other problems ; and therefore arc 
very ignorantly left out by Tacquet and Dechales in their 
editiv)ns of the Elements, who pretend that thev are scarce of 
any use : The cases of these problems, wherein it is required 
to apply a rectangle which shall be equal to a given square j 
to a given straig:it line, either deficient or exceeding by a 
square j are very often made use of by geometers : And, on this 
account, it is thought proper, for the sake of beginners, to give 
their constructions as follows : 

I. To apply a rectangle which shall be equal to a given 
square, to a given straight line, deficient by a square : But the 
given square must net be greater than that upon the half of the 
given line. 

Let AB be the given straight line, and let the square upon 
the given straight line C be that to which the rectangle to be 
applied must be equal, and this square by the determination is 
not greater than tHat upon half of the straight line AB. 

Bisedt AB in D, and if the square upon AD be equal to 
the square upon C, the thing required is done : But if it be not 
equal to it, AD must be 
greater than C, according 
to the determination : Draw 
DE at right angles to AB 
and make it equal to C ; 
produce ED to F, so that 
EF be equal to AD or DB, 
and from the centre E, at 
the distance EF, describe a 
circle meeting AB in G, 

and upon GB describe the square GBKH, and complete the 
re<Stangle AGHL ; also join EG : And because AB is bisected 
in D, the redtangle AG, GB together with the square of DG 
is equal* to (the square of DB, that is, of EF or EG, that is . 5 2 

Y4 to) ' ■ 







H K 


/-^ 


A 


J) 


/G B 


C 





328 NOTES. 

Book VI. to) the squares of ED, DG : Take away the square of DG 
'^^^'"^'^ from each of these equals ; therefore the remaining re£langle 
AG, GB is equal to the square of ED, that is, of C : But the 
rectangle AG, GP is the redlangle AH, because GH is equal 
to GB } therefore h ereftangle AH is equal to the given square 
' upon the straight une C. Wherefore the reftangle AH, equal 
to the given square upon C, has been applied to the given 
straight line AB, deficient by the square GK. Which was 
to be done. 

2. To apply a reftangle which shall be equal to a given 
square, to a given straight line, exceeding by a square. 

Let AB be the given straight line, a,nd let the square upon 
the given straight line C be that to which the redlangle to be 
applied must be equal. 

Bisect AB in D, and draw BE at right angles to it, so that 
BE be equal to C ; and having joined DE, from the centre D 
at the distance DE describe a circle meeting AB produced in 
G ; upon BG describe the square 
BGHK, and complete the rect- 
angle AGHL. And because AB 
is bisedled in D, and produced 
to G, the redlangle AG, GB 
together with the square of DB 
•6. 2. is equal* to (the square of DGj 
or DE, that is, to) the squares 
ofEB,BD. From each of these ^ 

equals take the square of DB ; 

therefore the remaining redfangle AG, GB is equal to the 
square of BE, that is, to the square upon C. But the rectan- 
gle AG,GB is the redtangle AH, because GH is equal toGB. 
Therefore the reCtangle AH is equal to the square upon C. 
Wherefore the rectangle AH, equal to the given square upon 
C, has been applied to the given straight line AB, exceeding 
by the square GK. Which was to be done. 

3. To apply a rectangle to a given straight line which shall 
be equal to a given reftangle, and be deficient by a square. 
l?ut the given redtangle must not be greater than the square 
upon the half of the given straight line. 

Let AB be the given straight line, and let the given reftan-, 
gle be that which is contained by the straight lines C, D, 
which is not greater than the square upon the half of AB ; it 
is required to apply to AB a rectangle equal to the redbngle 
C, D, deficient by a square. 

Draw 




i 



329 

Book VI. 



'3.3. 



" 6.2. 



N O T £ S. 

Draw AE, BF at right angles to AB, upon the same side of 
it, and make AE equal to C, and BF to D : join EF and bisedl 
it in G ; and from the centre G> at the distance Gii, describe a 
circle meeting AE again in H : Join HF, and draw GK pa- 
rallel to it, and GL parallel to AE, meeting AB in L. 

Because the angle EHF in a semicircle is equal to the right 
angle EAB, AB and HF are parallels, and AH and BF are 
parallels ; wherefore AH is equal to BF, and the reciangle 
EA, AH equal to the redangle EA, BF, that is, to the 
rediangle C, D : And because EG, GF are equal to one ano» 
ther, and AE, LG, BF parallels ; therefore AX- and LB are 
equal, also EK is equal to KH^ and the re6langle C, D, from 
the determination, is not greater than the square of AL, the 
half of AB ; wherefore the rectangle EA, AH is not greater 
than the square of AL, that is, of KG: Add to each the square 
of KE i therefore the square'' of AK is not greater than the 
squares of EK, KG, that is, 
than the square of EG ; and 
consequently the straight line 
AK or. GL is not greater 
than GE. Now, if GE be 
equal to GL, the circle EHF 
touches AB in L, and there- 
fore the square of AL is = 
equal to the rectangle EA, 
AH, that is, to the given reel- 
angle C, D : and that which 
was required is done : But 
if EG, GL be unequal, EG 
must be the greater : and 
therefore the circle EHF cuts the straight line AB : let it cut it 
in the points M, N, and upon NB describe the square NBOP, 
and complete the redangle ANPQ^: Because LM is equal to** 1 3 3 
LN, and it has been proved that AL is equal to LB ; there- 
fore A Vi is equal to XB, and the re£langle AN, NB equal to 
the redangle NA, AM, that is, to the rectangle "^ EA, AH, «Cor.36.S. 
or the rectangle C, D : But the rectangle AN, NB is the 
rectangle AP, because PN is equal to N B : Therefore the 
rectangle AP is equal to the rectangle C, D ; and the rectangle 
AP equal to the given rectangle C, D, has been applied to 
the given straight line AB, deficient by the square BP. 
Which was to be done. 

4.T^ 




«36. 3, 



330 

Book VI. 



'55. 3. 




NOTES. 

4. To apply a redangle to a given straight line that shall 
be equal to a given redangle, exceeding by a square. 

Let AB be the given straight line, and the rectangle C, D 
the given redangle, it is required to apply a redtangle to AB 
equal to C, D, exceeding by a square. 

Draw AE, BF at right angles to AB, on the contrary sides 
of it, and maice AE equal to C, and BF equal to D : Join 
EF, and bised it in G ; and from the centre G, at the distance 
GE, describe a circle meeting AE again in Hj join HF, and 
draw GL parallel to AE ; 
let the circle meet AB pro- 
duced in M, N, and upon 
BN describe the square 
BNOP, and complete the 
redlangle ANPQ_; because 
the angle EHF in a semicir- 
cle is equal to the right angle 
EAB, A3, and HF are pa- 
rallels, and therefore AH and 
BF are equal, and the rect- 
angle EA, AH equal to the 
redlangle EA, BF, that is, 

to the redangle C, D : And because ML is equal to LN, 
and AL to LB, therefore MA is equal to BN, and the reiStan- 
gle AN, NB to MA, AN, that is, ^ to the redangle EA, AH, 
or the rectangle C, D : Therefore the redangle AN, NB, 
that is, AP, is equal to the redangle C, D ; and to the given 
straight line AB the rectangle AP has bees applied equal to 
the given re(5tangle C, D, exceeding by the square BP. 
Which was to be done. 

Willebrordus Snellius was the first, as far as I know, who 
gave these constructions of the 3d and 4th Problems in his 
Apolionius Batavus : And afterwards the learned Dr. Halley 
gave them in the Scholium of the i8th Prop, of the 8th Book 
of ApoUonius's Conies restored by him. 

The 3d Problem is otherwise enunciated thus ; To cut a 
given straight line AB in the point N, so as to make the 
rectangle AN, NB, equal to a given space: Or, which is the 
same thing, having given AB the sum of the sides of a red- 
angle, and the magnitude of it being likewise given, to find its 
sides. 

And the 4th Problem is the same with this, To find a point 

N in the g^ven straight line AB produced, so as to make the 

3 rectangle 



NOTES. 331 

reftangle AN, NB equal to a given space : Or, which is the ^<»k VL 
same thing, having given AB the difference of the sides of a ^^'"'^^^ 
rectangle, and the magnitude of it, to find the sides. 

PROP. XXXI. B. VI. 

In the demonstration of this, the inversion of proportionals 
is twice negledlcd, and is now added, that the conclusion may 
be legitimately made by help of the 24.th Prop, of B. 5, as 
Clavius had done. 

PROP. XXXII. B. VI. 

The enunciation of the preceding 26th Prop, is not general 
enough ; because not only two similar parallelograms that have 
an angle common to both, are about the same diameter ; but 
likewise two similar parallelograms that have vertically oppo- 
site angles, have their diameters in the same straight lines : But 
there seems to have been another, and that a diredl demon- 
stration of these cases, to which this 32d Proposition was 
needful ; And the 32d may be otherwise, and something more 
briefly demonstrated, as follows : 

PROP. XXXII, B. VI. 

If two trianprles which have two sides of the one, Sec. 

Let GAF, HFC be two triangles which have two sides AG, 
GF, proportional to the two sides FH, HC, viz. AG to GF, 
as FH to HC ; and let AG be pa- 
rallel to FH, and GP to HC i AF 
and JFC are in a straight line. 

Draw CK parallel to FH, and 
Jet it meet GF produced in K; 
Because AG, KC are each of them 
parallel to FH, they are parallel'' 

to one another, and therefore the B^ -^7 -^ *3ai. 

alternate angles AGF, FKC are 

equal: And AG is to GF, as (FH to HC, that is=) CK to C34 j 
KF; wherefore the triangles AGF, CKF are equiangular'', < g ^ * 
and the angle AFG equal to the angle CKF : But GFK is a 
straight line, therefore AF and FC are in a straight line*. - • e^ ], 

The 26th Prop, is demonstrated from the 32d, as follows.; 

If two similar and similarly placed parallelograms have an 
angle common to both, or vertically opposite angles ; their 
diameters are in the game straight line. 

First, 




«3K I. 




•CQr.19.5, 



* 32. 6. 



NOTES. 

First, let the parallelograms ABCD, AEFG have the angle 
BAD common to both, and be similar, and similarly placed ; 
ABCDG, AEFG are about the same diameter. 

Produce EF, GF, to H, K, and join FA,FC ; then because 
the parallelograms ABCD, AEFG are similar, DA is to AB, 
as GA to AE : Wherefore the re- 
mainder DG is^ to the remainder 
EB, as GA to AE : But DG is 
equal toFH, EBtoFIC, andAE to 
CF : Therefore as FH to HC, so 
is AG to GF ; and FH, HC are 
parallel to AG, GF i and the tri- 
angles AGF, FHC are joined at 
one angle in the point F j where- 
fore AF, FC are in the same" straight line^ 

Next, Let the parallelograms KFHC, GFEA which are 
similar and similarly placed, have theii angles KFH, GFE 
vertically opposite ; their diameters AF, FC are in the same 
straight line. 

Because AG, GF are parallel to FH, HC; and that AG is 
to GF, as FH to HC; therefore AF, FC are in the same 
straight line''. 




PROP. XXXni. B. VI. 

The words " because they are at the centre," are left out 
as the addition of some unskilful hand. 

In the Greek, as also in the Latin translation, the words 
a. sTOX^, " any whatever," are left out in the demonstration of 
both parts of the proposition, and are now added as quite 
necessary ; and, in the demonstration of the second part, where 
the triangle BGC, is proved to be equal to CGK, the illative 
particle a.(,x^ in the Greek text ought to be omitted.' 

The second part of the proposition is an addition of Theon's, 
as he tells in his commentary on Ptolemy's MsyaXn 2t/vTa^»f, 
p. 50. 



PROP. B. C. D. B. VI. 



These three propositions are added, because they are fre- 
quently made use of by geometers. 

DEF. 



NOTES. 



DEF. IX and XL B. XL 

1 HE similitude of plane figures is defined from the equa- 
lity of their angles, and the proportionality of the sides about 
the equal angles ; for from the proportionality of the sides 
only, or only from the equality of the angles, the similitude 
of the figures does not follow, except in the case when the 
figures are triangles : The similar position of the sides which 
contain the figures, to one another, depending partly upon 
each of these : And for the same reason, those are similar 
solid figures which have all their solid angles equal, each 
to each, and are contained by the same number of similar 
plane figures : For there are some solid figures contained by 
similar plane figures, of the same number, and even of the 
same magnitude, that are neither similar nor equal, as shall be 
demonstrated after the notes on the lOth definition : Upon 
this account it was necessary to amend the definition of simi- 
lar solid figures, and so place the definition of a solid angle 
before it : And from this and the loth definition, it is suffi- 
ciently plain how much the Elements have been spoiled by 
unskilful editors. 

DEF. X. B. XL 

Since the meaning of the word "equal" is known and 
established before it comes to be used in this definition : 
therefore the proposition which is the loth definition of this 
book, is a theorem, the truth or falsehood of which ought to 
be demonstrated, not assumed ; so that Theon, or some 
other editor, has ignorantly turned a theorem, which ought 
to be demonstrated in this loth definition : That figures are 
similar, ought to be proved from the definition of similar> 
figures ; that they are equal, ought to be demonstrated from 
the axiom, "Magnitudes that wholly coincide are equal to 
" one another;" or from Prop. A. of Book 5, or the 9th 
Prop, or the 14th of the same Book, from one of which the 
equality of all kinds of figures must ultimately be deduced. In 
the preceding books, Euclid has given no definition of equal 
figures, and it is certain he did not give this : For what is 
called the first def. of the third book, is really a theorem in 
which these circles are said to be equal, that have the straight 
lines from the centres to the circumferences equal, which is 
plain, from the definition of a circle j and therefore has by" 

some 




334 



NOTES. 



12.11. 



Book XI. some editor been improperly placed among the definitions. 

^^''^"^^ The equality of figures ought not to be defined, but demon- 
strated ; Therefore, though it were true, that solid figures 
contained by the same number of similar and equal plane fi- 
gures are equal to one another, yet he would justly deserve to 
be blamed who would make a definition of this proposition, 
which ought to be demonstrated. But if this proposition be 
not true, must it not be confessed, that geometers have, for 
these thirteen hundred years, been mistaken in this elementa- 
ry matter? And this should teach us modesty, and to acknow- 
ledge how little, through the weakness of our minds, we are 
able to prevent mistakes, even in the principles of sciences 
which are justly reckoned amongst the most certain ; for that 
the proposition is not universally true, can be shewn by many 
examples: The following is sufficient: 

Let there be any plane redtilineal figure, as the triangle 
ABC, and from a point D within it draw* the straight Ime 
DE at right angles to the plane ABC ; in DE take DE, DF 
equal to one another, upon the opposite sides of the plane, 
and let G be any point in EF; join DA, DB, DC 5 EA, EB, 
EC ; FA, FB, FC ; GA, GB, GO : because the straight line 
EDF is at right angles to the plane ABC, it makes right an- 
gles with DA, DB, 'DC which it meets in that plane ; and in 
the triangles EDB, FDB, ED and DB are equal to FD and 1 
DB, each to each, and they contain right angles ; therefore 
the base EB is equal^ 
to the base FB ; in the 
same manner EA is 
equal to FA, and EC to 
FC: And in the triangles 
EBA, FBA, EB, BA, 
are equal to FB, BA> 
and the base EA is 
equal to the base FA ; 
wherefoi'c the angle 
EBA is equal " to the 
angle FBA, and the tri- 
angle EBA equal ^ to 
the triangle FBA, and 
the other angles equal to 
the other angles ; there- 
fore these triangles are 



»4.1. 



*^«. 1. 



Z^4. 6. 




similar ^ : In the ssme manner the triangle EBC is s/milar f 

the 






NOTES. 335 

the triatngle FBC, and the triangle EAC to FAC ; therefore ^o*"^ Xi. 
there are cwo solid figures, each of which is contained by six ^ 
triangles, one of them by three triangles, the common vertex 
of wnich is the point G, and their bases the straight lines AB, 
BC, C A, and by three other triangies, the common vertex of 
which is the point E, and their bases the same lines AB, 
BC, CA : The other solid is contained by the same three tri- 
angles the common vertex of which is G, and their bases AB, 
BC, CA : and by three other triangles of which the commoa 
vertex is the point F, and their bases the same straight lines 
AB, BC, CA : Nov^^ the three triangles GAB, GBC, GCA 
are common to both solids, and the t^ree others EAB, EBC, 
ECA^ ot the first solid, have been shewn equal and smiiar to the 
three others FAB, FBC, FC A of the other solid, each to each: 
therefore these two solids are contained by the sarre number of 
equal and similar planes : But that they are not equal is mani- 
fest, because the first of them is contained in the other : There- 
fore it is not universally true that solids are equal which are 
contained by the same number of equal and similar planes. 

Cor. From this it appears that two unequal solid angles may 
be contained by the same number of equal plane angles. 

For the solid angle at B, which is .contained by the four 
plane angles EBA, EBC, GBA, GBC is not equal to the 
solid angle at the same point B wnich is contained by the four 
plane angles FBA, FBC, GBA, GBC ; for this last contains 
the other : And each of them is contained hy four plane angles, 
which are equal to one another, each to each, or are the self 
same, as has been proved : And indeed there may be innu- 
merable solid angles all unequal to one another which are each 
of them contained by plane angles that are equal to one 
another, each to each: It is likewise manifest, that the before- 
mentioned solids are not similar, since their solid angles are 
not ail equal. 

And that there may be innumerable solid angles all unequal 
to one another, which are each of them contained by the same 
plane angles disposed in the same order, will be plain from 
the three following propositions. 



PROP. I. PROBLEM. 

Three magnitudes. A, B, C being given, to find a fourth 
«uch, that every three shall be greater than the remaining one. 

Let 



336 NOTES. 

Book XI. Let D be the fourth : therefore D must be less than A, B, 
^'^^'^'^^ C together : Of the three A, B, C, let A be that which is not 
less than either of the two B and C : And first, let B and C 
together be not less than A ; therefore B, C, D together are 
greater than A : and bacause A is not less than B j A, C, D 
together are greater than B : In the like manner A, B, D 
together are greater than C j Wherefore in the case in which 
B and C together are not less than A, any magnitude D which 
is less than A, B, C together, will answer the problem. 

But if B and C together be less than A ; then, because it is 
required that B, C, D together be greater than A, from each 
of these taking away B, C, the remaining one D must be 
greater than the excess of A above B and C : Take therefore 
any magnitude D which is less than A, B, C togeiher, but 
greater than the excess of A above B and C ; Then, B, C, D 
together are greater than A j and because A is greater than 
either B or C, much more will A and D, together with either 
of the two B, C be greater than the other : And, by the con- 
struction, A, B, C are together greater than D. 

Cor. If besides it be required, that A and B together shall 
not be less than C and D together ; the excess of A and B 
together above C must not be less than D, that is, D must 
not be greater than that excess. 



PROP. II. PROBLEM. 

Four magnitudes A, B, C, D being given, of which A and 
B together are not less than C and D together, and such that 
any three of them whatever' are greater than the fourth; it is 
required to find a fifth magnitude E such, that any two of the 
three A, B, E shall be greater than the third, and also that any 
two of the three C, D, E shall be greater than the third. Let 
A be not less than B, and C not less than D. 

First, Let the excess of C above D be not less than the 
excess of A above B : It is plain that a magnitude E can be 
taken which is less than the sum of C and D, but greater than 
the excess of C above D ; let it be taken ; then E is greater 
likewise than the excess of A above B ; wherefore E and B 
together are greater than A ; and A is not less than B ; there- 
fore A and E together are greater than B : And, by the hypo- 
tliesis, A and B together are not less than C and D together, 
and C and D together are greater than E > therefore likewise 
A and B are greater th:in E. - ' 

Bu^ 



NOTES 337 

But let the excess of A above B be greater than the excess Book xi. 
of C above D : And because, by the hypothesis, the three B, '"^^^'^^ 
C, Dare together greater than the fourth A ; C and D toge- 
ther are greater than the excess of A above B : Therefore a 
magnitude may be taken which is less than C and D together, 
but greater than the excess of A above B. Let this magnitude 
be E : and because E is greater than the excess of A above B, 
B together with E is greater than A : And, as in the preced- 
ing case, it may be shewn that A together with E is greater 
than B, and that A together with B is greater than E : There- 
fore, in each of the cases, it has been shewn, that any two of 
the three A, B, E are greater than the third. 

And because in each of the cases E is greater than the excess 
of C above D, E together with D is greater than C ; and by 
the hypothesis, C is not less than D ; therefore E together 
with C is greater than D ; and, by the construction, C and D 
together are greater than E ; Therefore any two of the three 
C, D, E are greater than the third. 

PROP. III. THEOREM. 

There may be innumerable solid angles all unequal to one 
another, each of which is contained by the same four plane 
angles, placed in the same order. 

Take three plane angles. A, B, C, of which A is not less 
than either of the other two, and such, that A and B toge- 
ther are less than two right angles: and, by Problem i, and 
its corollar}', find a fourth angle D such, that any three what- 
ever of the angles A, B, C, D be greater than the remaining 
anjgle, and su«h, that A and B together be not less than C 
and D together : And, by Problem 2, find a fifth angle E such, 
that any two of the angles A, B, E, be greater than the third, 




and also that any two of the angles C, D, E be greater than 

Z the 



338 



Book XI. 



NOTES. 

the third : And because A and B together are less than two 
right angles, the double of A and B together is less than four 
right angles : But A and B together are greater than the angle 
E ; wherefore the double of A, B together is greater than 
the three angles A, B, E together, which three are conse- 
quently less than, four right angles > and every two of the 
same angles, A, B, E are greater than the third ; therefore, by 
prop. 23, II, a solid angle may be made contained by three 
plane angles, equal to the angles A, B, E, each to each. Let 
this be the angle F, contained by the three plane angles GFH, 
HFK, GFK, which afe equal to the angles A, B, E, each to 
each : And because the angles C, D togerher are not greater 
than the angles A, B together, therefore the ju.gle? C, D, E 
are not greater than the angles A. B, E : But these last three 
^re less than four right angles, as has beendeuionstiated: where- 
fore also the angles C, 1), E are together less than four right 
angles, and every two of them are greatet than the third ; 
therefore a solid angle may be rriade, which shall be contained 
by three plane angles equal to the angles C, D, E, each to 




'!3. 11. 



each^ 1 And, by prop. 26, 11, at the point F, in the straight 
line FG, a solid angle may be made equal to that which is con- 
tained by the three plane angles that arc equal to the angles 
C, D, E : Let this be made, and let the angle GFK, which is 
equal to F', be one of the three ; and let KFL, GFL be the 
other two which are equal to the angles C, D, each to each. 
Thus there is a solid angle constituted at the point F, contained 
by the four plane angles GFH, HFK, KFL, GFL, which 
are equal to the angles A, B, C, D, each to each. 

Again, Find another angle M such, that every two of the 
three angles A, B. M be greater than the third, and also 
every two of the three C, D, M be greater than the third ; 

Ajvj, 




NOTES. 339 

And, as in the preceding part, it may be demonstrated, that ^^^l^- 
the three A, B, M are less 
than four right angles, as also 
that the three C, D, M, are 
less than four right angles. 

Make therefore^ a solid angle / \^ \^\^ »23. n. 

at N contained by the three 
plane angles ONP, PNQ_, 
ONQ_, which are equal to A, 
B, M, each to each : And by 
prop. 26, II, make at the 

point N, in the straight line ON, a solid angle contained by 
three plane angles, of which one is the angle ONQ equal toM, 
and the other two are the angles QNR, ONR, which are equal 
to the angles C, D, each to each. Thus, at the point N, there 
is a solid angle contained by the four plane angles ON P, PNQ, 
QNR, ONR which are equal to the angles A, B, C, D, each 
to each. And that the two solid angles at the points P, N, 
each of which is contained by the above-named four plane 
angles, are not equal to one another, or that they cannot coin- 
cide, will be plain by considering that the angles GFK, ONQ : 
that is, the angles E, M, are unequal by the construction ; and 
therefore the straight lines GF, FK cannot coincide with ON, 
NQ, nor consequently can the solid angles, which therefore 
are unequal. 

And because from the four plane angles A, B, C, D, there 
can be found innumerable other angles that will serve the same 
purpose with the angles E and M : it is plain that innumerable 
other solid angles may be constituted which are each contained 
by the same four plane angles, and all of them unequal to one 
another. Q^ E. D. 

And from this it appears, that Clavius and other authors are 
mistaken, who assert that those solid angles are equal which 
are contained by the same number of plane angles that are 
equal to one another, each to each. Also it is plain, that the 
26th prop, of Book 1 1, is by no means sufficiently demonstrat- 
ed, because the equality of two solid angles, whereof each is 
contained by three plane angles which are equal to one another, 
tach to each, is only assumed, and not demonstrated. 

Z 2 



340 

r > K XI. 



NOTES. 

PROP. I. B. XI. 

The words at the end of this, *' for a straight line cannot 
*' meet a straight line in more than one point," are left out, as 
an addition by some unskilful hand ; for this is to be demon- 
strated, not assumed. 

Mr. Thomas Simpson, in his notes at the end of the second 
edition of his Elements of Geometry, p. 262, after repeating 
the words of this note, adds " Now, cata it possibly shew any 
" want of skill in an editor (he means Euclid or Theon) tore- 
<' fer to an axiom which Euclid himselfhath laid down, Book i, 
** N° 14," he means Barrow's Euclid, for it is the loth in the 
Greek, *' and not to have demonstrated, what no man can 
** demonstrate ?" But all that in this case can follow from that 
axiom is, that, if two straight lines could meet each other in 
two points, the parts of them betwixt these points must coin- 
cide, and so they would have a segment betwixt these points 
common to both. Now, as it has not been shewn in Euclid, 
that they cannot have a common segment, this does not prove 
that they cannot meet in two points, from which their not 
having a common segment is deduced in the Greek edition : 
But, on the contrary, because they cannot have a common seg- 
ment, as is shewn in Cor. of iith Prop. Book i, of 4to edi- 
tion, it follows plainly, that they cannot meet in two points, 
which the remarker says no man can demonstrate. 

Mr. Simpson, in the same notes, p. 265, justly observes, that 
in the corollary of Prop. 1 1, Book i, 410 edit, the straight lines 
AB, BD, BC, are supposed to be all in the same plane, which 
cannot beassumed in ist Prop. Book 1 1. This, soon after the 
4to edition was published, I observed and corrected as it is 
now in tbis edition : He is mistaken in thinking the loth 
axiom he mentions here to be Euclid's ; it is none of Euclid's, 
but is the loth in Dr. Barrow's edition, who had it from He- 
rigon's Cursus, vol. I. and in place of it the corollary of nth 
Prop. Book I, was added. 

PROP. II. B. XI. 

This proposition seems to have been changed and vitiated 
by some editor ; for all the figures defined in the ist Book of 
the Elements, and among them triangles, are, by the hypo- 
thesis, plane figures ; that is, such as are described in a plane ; 
wherefore the second part of the enunciation needs no demon- 
stration. Besides, a convex superficies may be terminated by 

th|:ee 



NOTES. 341 

three straight lines meeting one another : The thing that ^°°|^5^ 
should have been demonstrated is, that two or three straight ""^^^^"""^ 
lines, that meet one another, are in one plane. And as this 
is not sufficiently done, the enunciation and demonstration are 
changed into those now put into the text. 

PROP. III. B. XI. 

In this proposition the following words near to the end of it 
are left out, viz. " therefore DEB, DFB are not straight lines ; 
" in the liice manner it may be demonstrated, that there can be 
*' no other straight line between the points D, B ;" Because 
from this, that two lines include a space, it only follows that 
one of them is not a straight line : And the force of the argu» 
raent lies in this, viz. if the common section of the planes be 
not a straight line, then two straight lines could include a space, 
which is absurd j therefore the common section is a straight line. 

PROP. IV. B. XI. 

The words " and the triangle AED to the triangle BEC* 
are omitted, because the whole conclusion of the 4th Prop. 
B. I, has been so often repeated in the preceding books, it was 
needless to repeat it here. 

PROP. V. B. XI. 

In this, near to the end, ttiinrt^u ought to be left out in the 
Greek text : And the word " plane" is rightly left out in the 
Oxford edition of Commandine's translation. 

PROP. Vll. B. XI. 

This proposition has been put into this book by some un» 
skilful editor, as is evident from this, that straight lines which 
are drawn from one point to another in a plane, are, in the 
preceding books, supposed to be in that plane : And if they 
were not, some demonstrations in which one straight line is 
supposed to meet another would not be conclusive, because 
these lines would not meet one another : For instance, in Prop. 
30, B. I, the straight line GK would not meet EF, if GK. were 
not in the plane in which are the parallels AB, CD, and in 
which, by hypothesis, the straight line EF is : Besides, this 7th 
Proposition is demonstrated by the preceding 3d ; in which the 
very thing which is proposed to be demonstrated in the 7th 
is twice assumed, viz. that the straight line drawn from one 
point to another in a plane, is in that plane ; and the same thing 
IS assumed in the preceding 6thProjx in which the straight line 

Z 3 which 



342 NOTES. 



Book XI. 



which joins the points B,- D that are in the plane to which 
AB and CD are at right angles, is supposed to be in that 
plane : And the 7th, of which another demonstration is given, 
is kept in the book merely to preserve the number of the pro- 
positions ; for it is evident, from the yth and 35th definitions 
of the ist book, though it had not been in the Elements. 

PROP. VIII. B. XL 

In the Greek, and in Commandine's and Dr. Gregory's 
translations, near to the end of this Proposition, are the fol- 
lowing words : " But DC is in the plane through BA, AD," 
instead of which, in the Oxford edition of Commandine's trans- 
lation is rightly put, *' but DC is in the plane through BD, 
DA." But all the editions have the tollowing words, viz. 
" because A, B, BD are in the plane through BD, DA, and 
" DC is in the plane in which are AB, BD," which are ma- 
nifestly corrupted, or have been added to the text ; for there 
was not the least necessity to go so far about to shew that DC 
is in the same plane in which are BD, DA, because it imme- 
diately follows, from Prop. 7, preceding, that BD, DA, are 
in the plane in which are the parallels AB, CD : I'herefore, 
instead of these words, there ought only to be, " because all 
<' three are in the plane in which are the parallels AB, CD.'* 

PROP. XV. B. XI. 

After the words, " and because BA is parallel to GH," 
the following are added, *' for each of them is parallel to DE, 
" and are not both in the same plane with it," as being mani- 
festly forgotten to be put into the text. 

PROP. XVI. B. XI. 

In this, near to the end, instead of the words, " but straight 
" lines which meet neither way," ought^o be read, " but 
" straight lines in the same plane which produced meet neither 
" way :" Because, though in citing this definition in Prop. 27, 
Book I, it was not necessary to mention the words, " in the 
*■*■ same plane," all the straight lines in the books preceding 
this being in the same plane; yet here it was quite necessary. 

PROP. XX. B. XL 

In this, near the beginning, are the words, " But if not, let 
" BAC be the greater :" But the anjjle BAG may happen to 
Ve equal to one of the other two : Wherefore this place should 

be 



I 



NOTES. 



343 



be read thus, " But if not, let the angle BAC be not lessthan BoosXl. 
" either of the other two, but greater than DAB." v<^<-"*^ 

At the end of this proposition it is said, " in the same man- 
*' ner it may be demonstrated," though there is no need of any 
demonstration j because the angle BAC being not less than 
either of the other two, it is evident that BAC together with 
one of them is greater than the other. 

PROP. XXII. B. XI. 
And likewise in this, near the beginning, it is said, " But 
** if not, let the angles at B, E, H be unequal, and let the 
*' angle at B be greater than either of those at E, H ;" Whichi 
words manifestly shew this place to be vitiated, because the 
angle at B may be equal to one of the other two. They ou^ht 
therefore to be read thus, " But if not, let the angles at B, E, 
" H be unequal, and let the angle at B be not less than either 
" of the other two at E, H : Therefore the straight line AC 
" is not less than either of the two DF, GK." 

PROP. XXIII. B. XI. 

The demonstration of this is made something shorter, by not 
repeating in the third case the things which were demonstrated 
in the first ; and by making use of the construction which 
Campanus has given ; but he does not demonstrate the second 
and third cases : The construdlion and demonstration of the 
third case are made a little more simple than in the Greek text. 

PROP. XXIV. B. XI. 

The word *' similar" is added to the enunciation of this pro- 
position, because the planes containing the solids which are to 
be demonstrated to be equal to one another, in the 25th pro- 
position, ought to be similar and equal, that the equality of the 
solids may be inferred from Prop. C. of this Book; And in the 
Oxford edition of Commandine's translation, a corollary is 
added to Prop. 24, to shew that the parallelograms mentioned 
in this proposition are similar, that the equality of the solids in 
Prop. 25, may be deduced from the loth def. of Book ii. 

PROP. XXV and XXVI. B. XI. 

In the 25th Prop, solid figures which are contained by the 
same number of similar and equal plane figures, are supposed 
to be equal to one another. And it seems that Theon, or some 

Z 4 other 



344 NOTES. 

Book xr. Other editor, that he might save himself the trouble of de- 

'^^^'^^ monstrating the solid figures mentioned in this proposition to 

be equal to one another, has inserted the loth def. of this 

Book, to serve instead of a demonstration: which was very 

ignorantly done. 

Likewise in the 26th Prop, two solid angles are supposed to 
be equal : If each of them be contained by three plane angles 
which are equal to one another, each to each. And it is strange 
enough, that none of the commentators on Euclid have, as far 
as I know, perceived, that something is wanting in the demon- 
strations of these two propositions. Clavius, indeed, in a note 
upon the nth def, of this Book, affirms, that it is evident that 
those solid angles are equal which are contained by the same 
number of plane angles, equal to one another, each to each, 
because they will coincide, if they be conceived to be placed 
within one another ; but this is said without any proof, nor is it 
always true, except when the solid angles are contained by three 
plane angles only, which are equal to one another, each to 
each : And in this case the proposition is the same with this, 
that two spherical triangles that are equilateral to one another, 
are also equiangular to one another, and can coincide : which 
ought not to be granted without a demonstration, Euclid does 
not assume this in the case of reftilineal triangles, but demon- 
strates in Prop. 8, Book i, that triangles which are equilateral 
to one another, are also equiangular to one another ; and from 
this their total equality appears by Prop. 4. Book i. AndMe- 
nelaus, in the 4th Prop, of his first Book of Spherics, explicitly 
demonstrates, that spherical triangles which are mutually equi- 
lateral, are also equiangular to one another ; from which it is 
easy to shew that they must coincide, providing they have 
their sides disposed in the same order and situation. 

To supply these defeats, it was necessary to add the three 
Propositions marked A, B, C to this Book. For the 25th, 
26th, and 28th Propositions of it,and consequently eight others, 
viz. the 27th, 31st, 32d, 33d, 34th, 36th, 37th, and 40th of 
the same, which depend upon them, have hitherto stood upon 
an infirm foundation j a« also, the 8th, 12th Cor. of 17th and 
18th of I2th Book, which depend upon the 9th definition. For 
it has been shewn in the notes on def. loth of this book, that 
solid figures which are contained by the same number of simi- 
lar and equal plane figures, as also solid angles that are con- 
tained by the same number of equal plaac angles, are not 
always equal to one another. 

It 



I 



NOTES. 3^5 

Ic is to be observed that Tacquet, in his Euclid, defines Book XI. 
equal solid angles tobe such, " as being put within one another ^"^^v^*^ 
« do coincide ;" but this is an axiom, not a definition ; for it is 
true of all magnitudes whatever. He made this useless defini- 
tion, that by it he might demonstrate the 36th Prop, of this 
iSook, without th? help of the 35th of the same : Concerning 
which demonstration, see the note upon Prop. 36. 

PROP. XXVIII. B. xr. 

In this it ought to have been demonstrated, not assumed, 
that the diagonals are in one plane. Clavius has supplied this 
defect. 

PROP. XXIX. B. XI. 

There are three cases of this proposition; the first is, when 
the two parallelograms opposite to the base AB have a side 
common to both : the second is, when these parallelograms are 
separated from one another ; and the third, when there is a 
part of them common to both ; and to this last only, the de- 
monstration that has hitherto been in the Elements does agree. 
The first case is immediately deduced from the preceding 28th 
Prop, which seems for this purpose to have been premised to 
this 29ch, for it is necessary to none but to rt, and to the 40th 
of this book, as we now have it, to which last, it would, with- 
out doubt, have been premised, if Euclid had net made use of 
it in the 29th ; but some unskilful editor has taken it away 
from the Elements, and has mutilated Euclid's demonstratioa 
of the other two cases, which is now restored, and serves for 

both at once. 

» 

PROP. XXX. B. XL 

In the demonstration of this, the opposite planes of the solid 
CP, in the figure in this edition, that is, of the solid CO in 
Commandine's figure, are not proved to be parallel ; which it 
is proper to do for the sake of learners. 

PROP. XXXI. B. XI. 

There are two cases of this proposition ; the first is, when 
the insisting straight lines are at right angles to the bases j the 
other, when they are not ; the first case is divided again into 
two others, one of which is, when the bases are equiangular 
parallelograms ; the other, when they are not equiangular : 

The 



346 NOTES. 

Book xr. The Greek editor makes no mention of the first of these two 
"^^^^ last cases, but has inserted the demonstration of it as a part of 
that of the other : And therefore should have taken notice of 
it ill a corollary ; but we thought it better to give these two 
cases separately : The demonstration also is made something 
shorter by following the way Euclid has made use of in Prop. 
14, Book 6. Besides, in the demonstration of the case in which 
the insisting straight lines are not at right angles to the bases, 
the editor does not prove that the solids described in the con- 
struftion, are parallelopipeds, which it is not to be thought 
that Euclid neglected : also the words, " of which the insistmg 
*' straight lines are not in the same straight lines," have been 
added by some unskilful hand ; for they may be in the'same 
straight lines. 

PROP. XXXII. B. XI. 

The editor has forgot to order the parallelogram FH to be 
applied in the angle FGH equal to the angle LCG, which is 
necessary. Ciavius has supplied this. 

Also, in the const rudion, it is required to complete the 
solid of which the base is FH, and altitude the same with that 
of the solid CD : But this does not determine the solid to be 
completed, since there may be innumerable solids upon the 
same base, and of the same altitude : It ought therefore to be 
said, " complete the solid of which the base is FH, and one of 
" its insisting straight lines is FD :" The same corredtion 
must be made in the following Proposition 33. 

PROP. D. B. XI. 

It is very probable that 'Euclid gave this proposition a place 
in the Elements, since he gave the like proposition concerning 
equiangula: parallelograms in the 23d B. 6. 

PROP. XXXIV. B. XI. 

In thiS the words, uv xt t^to-rwo-ai HKiia-iv Inji rZv avrwv tvOttZ*. 
" of which the insisting straight lines are not in the same 
" straight lines," are thrice repeated : but these words ought 
either to be left out, as they are byClaviu;^, or, in place of 
them, ought to be put, " whether the insisting straight lines be, 
*' or be not in the same straight lines :" For the other case is 
without any reason, excluded j also the words Hv ra. v^^n, c 

" whic 



1 



NO T E S. 347 

** which the altitudes," are twice put for ;»ii l^trrxirxt, " of Book XL 
** which the insisting straight lines ;'* which is a plam mistake . 
For the altitude is always at right angles to the base. 

PROP. XXXV. B. XI. 

The angles ABH, BEM are demonstrated to be right an- 
gles in a shorter way than in the Greek ; and in the same way 
ACH, DFM may be demonstrated to be right angles : Also 
the repetition ot the same demonstration, which begins with 
" in the same manner," is lett out, as it was probably added 
to the text by some editor : for the words " in like manner 
*' we mav demonstrate," are not inserted except when the 
demonstration is net given, or when it is something ditterent 
from the other if it be given, as in Prop. 26, of this Book. 
Campatms has not this repetition. 

We have given another demonstration of the corollary, 
besides the one in the original, by help of which the following 
36ch Prop, may be demonstrated without the 35th. 

PROP. XXX VT B. XI. 

T ACQUfiT in his Euclid demonstrates this proposition with- 
out the help of the 35th ; but it is plain, that the solids men- 
tioned in the Greek text in the enunciation of the proposiyon 
as equiangular, are such that their solid angles are contained 
by three plane angles equal to one another, each to each; as is 
evident from the construction, ^ow Tacquet does not de- 
monstrate, but assumes these solid angles to be equal to one 
another ; for he supposes the solids to be already made, and 
does not give the construction by which they are made : But, . 
by the second demonstration of the preceding corollary, his 
demonstration is rendered legitimate likewise in the case where 
the solids are constructed as in the text. 

PROP. XXXVII. B. XL 

In this it is assuraeH, that the ratios which are triplicate of 
those ratios which are the same with one another, are likewise 
the same with one another ; and that those ratios are the same 
with one another, of which the triplicate ratios are the same 
with one another ; but this ought not to be granted without a 
demonstration ; n.K did Euclid assume the first and easiest of 
these two propositions, but demonstrated it in the case of 
duplicate ratios, in the 22nd Prop. Book 6. On this account, 
another demonstration is given of this Proposition like to that 
vhich Euclid gives in Prop. 22, Bock 6, as Clavius has done. 




• 17. n. 



NOTES. 

PROP. XXXVIII. B. XI. 

When it is required to draw a perpendicular from a point 
in one plane which is at right angles to another plane, unto 
this last plane, it is done by drawing a perpendicular from the 
point to the common section of the planes ; for this perpen- 
dicular will be perpendicular to the plane, by def. 4, of this 
Book : And it would be foolish in this case to do it by the nth 
Prop, of the same ; But Euclid% Apollonius, and other geo- 
in'otWr meters, when they have occasion for this problem, direct a 
editiohs. perpendicular to be drawn from the point to the plane, and 
conclude that it will fall upon the common section oftheplanes, 
because this is the very same thing as if they had made use of 
the construction above mentioned, and then concluded that th€ 
, straight line must be perpendicular to the plane : but is ex- 
pressed in fewer words : Some editor, not perceiving this, 
thought it was necessary to add this proposition, which can 
never be of any use to the i ith book, and its being near to the 
end among propositions with which it has no connexion, is a 
mark of its having been added to the text. 

PROP. XXXiX. B. XI. 

In this it is supposed, that the straight lines which bisect 
the sides of the opposite planes, are in one plane, which ought 
to have been demonstrated ; as is now done, 

B. XII. 

B»oK XII. JL HE learned Mr. Moore, Professor of Greek in the Uni- 
^'^'^^"^ rersity of Glasgow, observed to me, that it plainly appears 
from Archimedes's Epistle to Dositheus, prefixed to his books 
of the Sphere and Cylinder, which epistle he has restored from 
ancient manuscripts, that Eudoxus was the author of the chief 
propositions in this 12th book. 

PROP. II. B. XII. 

At the beginning of this it is said, " if it be not so, the square 

*< of BD shall be to the square of FH, as the circle ABCD is 

( " to some space either less than the circle EFGH, or greater 

" than it :" And the like is to be found near to the end of this 

proposition, as also in Prop. 5, 11, 12, 18, of this Book: Con- 

3 cerning 



it P T E S. 34.9 

«ej-ning which it is to be observed, that in the demonstration Book XII. 
of theorems, it is sufficient, in this and the like cases, that a ^-^v^^ 
thing made use of in the reasoning can possibly exist, provid- 
ing this be evident, though it cannot be exhibited or found 
by a geometrical constru«£tion : So, in this place, it is assumed, 
that there may be a fourth proportional to these three magni- 
tudes, viz. the squares of BD, FH, and the circle ABCD j 
because it is evident that there is some square equal to the cir- 
cle ABCD though it cannot be found geometrically; and to 
the three rectilineal figures, viz. the squares of BD, FH, and 
the square which is equal to the circle ABCD, there is a fourth 
square proportional j because to the three straight lines which 
are their sides, there is a fourth straight line proportional*, « 12. t 
and this fourth square, or a space equal to it, is the space which 
in this proposition is denoted by the letter S : And the like is to 
be under stood in the other places above cited: And it is pro- 
bable that this has been shewn by Euclid, but left out bysome 
editor ; for the lemma which some unskilful hand has ^yided 
to this proposition explains nothing of it. 

PROP. III. B. XII. 

In the Greek text and the translations, it is said, <* and 
because the two straight lines BA, AC which meet one ano- 
ther," &c. here the angles BAC, KHL, are demonstrated 
to be equal to one another, by loth Prop. B. 11, which had 
been done before : Because the triangle EAG was proved to ' 
be similar to the triangle KHL : This repetition is left out, 
and the triangles BaC, KHL are proved to be similar in a 
shorter way by Prop. 21. B. 6. 

PROP. IV. B. XII. 

A FEW things in this are more fully explained, than in the 
Greek text. 

PROP. V. B. XIL 

Ik this, near to the end, are the words, is t (Av§o<r^tr thi^lh, 
as was before shewn ;" and the same are found again in the \ 

end of Prop. 18, of this Book; but the demonstration -referred 
io, except it be in the useless lemma annexed to the 2d Prop. 
i« no where in these Elements, and has bctn perhaps left out 
►f some editor, who has forgot to cancel those words al»o. 




N O T E S. 

PROP.Vr. B.Xil. 

A SHORTER demonstration is given of this; and that which 
is in the Greek text may be made shorter by a step than it is : 
For the author of it makes use of the 22d Prop, of B, 5, twice : 
Whereas once would have served his purpose ; because that 
proposition extends to any number of magnitudes which are 
proportionals taken two and two, as well as to three which 
are proportional to other three, 

COR. PROP. VIII. B. XII. 

The demonstration of this is imperfe£b, because it is nqt 
shewn, that the triangular pyraniids into which those upon 
multangular bases are divided, are similar to one another, as 
ought necessarily to have been done, and is done in the like 
case in Prop. 12 of this Book : The full demonstration of the 
corollary is as follows : 

Upon the polygonal bases ABCDE, FGHKL, let there be 
similar and similarly situated pyramids which have the points 
M, N for their vertices: The pyramid ABCDEM has to the 
pyramid FGHKLN the triplicate ratio of that which the side 
AB has to the homologous side FG. 

Let the polygons be divided into the triangles ABE, EBC, 

» 20. 6. ECD J FGL, LGH, LHK, which are similar^ each to each : 

"li.def.n And because the pyramids are similar, therefore"^ the triangle 

EAM is similar to the triangle LFN, and the triangle ABM 

' 4. &. to FGN : Wherefore^ ME is to E A, as NL to LF j and a 




AE to EB, so is FL to LG, b.cause the triangk-s EAB, 1 ,FG 
are similar ; therefore, ex aequali, as ME t© EB, so is NL tc 

LQ; 



N O T E S. 351 

LG : In like manner it may be shewn, that EB is to BM, as ^"'"'^ ^i^- 
LGto GN ; therefore, again, ex aequali, as EM to MB, so is ""'^ 
LN to NG : Wherefore the triangles EMB, LNG having 
their sides proportionals, are"^ equiangular, and similar to one '^ 5. 6. 
another : Therefore the pyramids which have the triangles 
EAB, LFG for their bases,'and the points M, N for their ver- 
tices, are similar^ to one another, for their solid angles are*=, oii.def.a 
equal, and the solids themselves are contained by the same num- ' B- li- 
ber of similar planes: In the same manner the pyramid EBCM 
may be shewn to be similar to the pyramid LGHN, and the 
pyramid ECDM to LHKN : And because the pyramids 
EABM, LFGN are similar, and have triangular bases, the py- 
ramid EABMhas*^ to LFGN the triplicate ratioof that which 'a. 12. 
EB has to the homologous side LG. And, in the same man- 
ner, the pyramid EBCM has to the pyramid LGHN the 
triplicate ratio of that which EB has to LG : Therefore as the 
pyramid EABM is to the pyramid LFGN, so is the pyramid 
EBCM to the pyramid LGHN : In like manner, as the pyra^ 
mid EBCM is to LGHN, so is the pyramid ECDM to the 
pyramid LHKN : And as one of the antecedents is to one of 
the consequents, so are all the antecedents to all the conse- 
quents : Therefore as the pvramid EABM to the pyramid 
LFGN, so is the whole pyramid ABCDEM to the whole 
pyramid FGHKLN : And the pyramid EABM has to the 
pyramid LFGN the triplicate ratio of that which AB has to 
EG ; therefore the whole pyramid has to the whole pyramid 
the triplicate ratio of that which AB has to tke homologous 
side FG. Q. E. D. 

PROP. XI and XIL B. XIL 

The order of the letters of the alphabet is not observed in 
these two propositions, according to Euclid's manner, and is 
now restored : By which means, the first part of prop: 12, may 
be demonstrated in the same words with the first part of Prop. 
II.; on this account the demonstration of that first part is 
left out and assumed from Prop. ii. 

PROP. XIII. B. XIL 

In this proposition, the common ssflion of a plane parallel 
to the bases of a cylinder, with the cylinder itself, is supposed 
to be a circle, and it was thought proper briefly to demon- 
strate it -, from whence it is sufficiently manifest, that this 
plane divides the cylinder into two others : And the same thing 
is understood to be suppjied in Prop. 14. 




K o r fi s. 



PROP. XV. B. XII. 

** And complete the cylinders AX, EO," both the enun- 
ciation and exposition of the proposition represent the cylin- 
ders as well as the cones, as already described : Wherefore 
the reading ought rather to be, " and let the cones be ALC, 
" ENG i and the cylinders AX, EO." 

The first case in the second part of the demonstration is 
wanting ; and something also in the second case of that part, 
before the repetition of the constru<Siion is mentioned j which 
are now added. 



PROP. XVII. B. XII. 
In the enunciation of this proposition, the Greek word* . 

tu TJiv (xii^ovx a^atK^OLv am^isv C70>tt5fov ty[^a,<^ai fxi xjayoj tas: tKxv- 
vovos tr^at^itt nxrat. t»j» iittpttniuv are thus translated by Comman- 
dine and others, " in majori solidum polyhedrum describerc 
'*'^ quod minoris sphseras superficiem noii tangat i" that is, " to 
** describe in the greater sphere a solid polyhedron which shall 
" not meet the superficies of the lesser sphere :" Whereby 
they refer the words >«»T(Z t%v tTtifanta.-: to these next to them 
ms iyx<yc6\os a^xifixs: But they ought by no means to be thus 
translated; for the solid polyhedron doth not only meet the 
superficies of the lesser sphere, but pervades the whole of that 
sphere : Therefore the aforesaid words are to be referred to 
TO e-Ti^fov TroXvtl^ov^ and Ought thus to be translated, viz, to 
describe in the greater sphere a solid polyhedron whose super- 
ficies shall not meet the lesser sphere; as the meaning of the 
proposition necessarily requires. 

The demonytratioh of the proposition is spoiled and muti- 
lafed : For some easy things are very explicitly demonstrated, 
while others not so obvious are not sufficiently explained ; for 
example, when it is affirmed, that the square of KB is greater 
than the double of the square of BZ, in the first demonstra- 
tion ; and that the angle BZK is obtuse, in the second : Both 
which ought to have been demonstrated : Besides, in the first 
demonstration, it is said, "draw Kfi from the point K, perpen- 
** dicular to BD;" whereas it ought to have been said, *' join 
** KV," and it should have been demonstrated, that KV is 
perpendicular to BD: For it is evident from the figure in Her- 
Y^gius's and Gregory's editions, and from the words of the 

demon- 



I 



NOTES. 3S3 

den^onstration, that the Greek editor did not perceive that the Book xil. 
perpendicular drawn from the point K to the straight line BD ^-^v^^ 
must necessarily fall upon the point V, for in the figure it is 
made to fall upon the point fi, a different point from V, which 
is likewise supposed in the demonstration. Commandine 
seems to have been aware of this ; for in this figure he marks 
one -and the same point with two letters V, D. ; and before 
Commandine, the learned John Dee, in the commentary he 
.annexes to this proposition in Henry Rillingsley's translation of 
the Elements, printed at London, ann. 1570, expressly takes 
notice of this error, and gives a demonstration suited to the 
construction in the Greek text, by which he shews that the 
perpendicular drawn from the point K to BD, must necessarily 
fall upon the point V. 

Likewise it is not demonstrated, that the quadrilateral figures 
SOFT, TPRY, and the triangle YRX, do not meet the les- 
ser sphere, as was necessary to have been done : only Clavius, 
as far as I know, has observed this, and demonstrated it by a 
lemma, which is now premised to this proposition, something 
altered, and more briefly demonstrated. 

In a corollary of this proposition, it is supposed that a solid 
polyhedron is described in the other sphere similar to that which 
is described in the sphere BCDE ; but, as the construdtion by 
which this may be done is not given, it was thought proper to 
give it, and to demonstrate, that the pyramids in it are similar 
to those of the same order in the solid polyhedron described 
in the sphere BCDE. 

From the preceding notes, it is sufficiently evident how 
much the Elements of Euclid, who was a most accurate geo- 
meter, have been vitiated and mutilated by ignorant editors. 
The opinion which the greatest part of learned men have en- 
tertained concerning the present Greek edition, viz. that it is 
very little or nothing different from the genuine work of 
Euclid, has without doubt deceived them, and made them less 
attentive and accurate in examining that edition ; whereby 
several errors, some of them gross enough, have escaped their 
notice from the age in which Theon lived to this time. Upon 
which account there is some ground to hope that the pains we 
have taken in correcting those errors, and freeing the Ele- 
ments as far as we could from blemishes, will not be unac- 
ceptable to good judges, who can discern when demonstrations 
are legitimate, and when they are not. 

A a The 



354 NOTES. 

BooKXit. The obje£lions which, since the first edition, have been 
^'"''''^'^ made against some things in the notes, especially against the 
doctrine of proportionals, have either been fully answered in 
Dr. Barrow's Le6l. Mathemat. and in these nates ; or are such, 
except one which has been taken notice of in the note on 
Prop. I. Book II, as shew that the person who made them 
has not sufficiently considered the things against which they 
are brought; so that it is not necessary to make any further 
answer to these objections and others like them against Euclid's 
definition of proportionals, of which definition Dr. Barrow 
justly says in page 297 of the above named book, that 
** Nisi machinls impulsa validioribus iEternum persistet incon-- 
*' cussa." 



END OF THE NOTE a. 



E U C L I D'S 

DATA, 



IN THIS EDITION 



"SEVERAL ERRORS ARE CORRECTED, 



AND 



SOME PROPOSITIONS ADDED. 



By ROBERT SIMSON, M. D. 

Emeritus Professor of Mathematics in the Unitersitj of Glasgow. 



LONDON^ 



Prifitcd for F. Wisgbave, in the Strand, Snccefsor to Mr. Nottrsc. 
1806. 



I 



PREFACE. 



EUCLID'S DATA is the first in order of the 
books written by the ancient geometers to facihtate 
and promote the method of resolution or analysis. 
In the general, a thing is said to be given which is 
either actually exhibited, or can be found out, that 
is, which is either known by hypothesis, or that can 
be demonstrated to be known ; and the propositions 
in the book of Euclid's Data shew what things can 
be found out or known from those that by hypo- 
thesis are already known ; so that in the analysis or 
investigation of a problem, from the things that are 
laid down to be known or given, by the help of 
these propositions other things are demonstrated to 
be given, and from these, other things are again 
shewn to be given, and so on, until that which was 
proposed to be found out in the problem is demon- 
strated to be given ; and when this is done, the 
problem is solved, and its composition is made and 
derived from the compositions of the Data which 
were made use of in the analysis. And thus the 
Data of Euclid are of the most general and ne- 
cessary use in the solution of problems of every 
kind. 

Euclid is reckoned to be the author of the 
Book of the Data, both by the ancient and modem 
geometers ; and there seems to be no doubt of his 
having written a book on this subject, but which, 
in the course of so many ages, has been much vi- 
tiated by unskilful editors in several places, both in 
the order of the propositions, and in the definitions 
and demonstrations themselves. To correct the 

A a 3 errors 



358 PREFACE. 

errors which are now found in it, and bring it 
nearer to the accuracy with which it was, no doubt, 
at first written by EucUd, is the design of this 
edition, that so it may be rendered more useful to 
geometers, at least to beginners who desire to learn 
the investigatory method of the ancients. And for 
their sakes, the compositions of most of the Data 
are subjoined to their demonstrations, that the 
compositions of problems solved by help of the 
Data may be the more easily made. 

Marin us the philosopher's preface, which, in 
the Greek edition, is prefixed to the Data, is here 
left out, as being of no use to understand them. 
At the end of it, he says, that Euclid has not used 
the synthetical but the analytical method in de- 
livering them ; in which he is quite mistaken ; for 
in the analysis of a theorem, the thing to be de- 
monstrated is assumed in the analysis ; but in the 
demonstrations of the Data, the thing to be de- 
monstrated, which is, that something or other is 
given, is never once assumed in the demonstration, 
from which it is manifest, that every one of them 
is demonstrated synthetically ; though indeed, if a 
proposition of the Data be turned into a problem, 
for example, the 84th or 85th in the former edi- 
tions, which here are the 85th and S6th, the de- 
monstration of the proposition becomes the analysis 
of the problem. 

Wherein this edition differs from the Greek, 
and the reasons of the alterations from it, will be 
shewn in the notes at the end of the Data. 



EUCLID S DATA. 



DEFINITIONS. 



i^PACES, lines, and angles, are said to be given in magni- 
tude, when equals to them can be found. 

II. 
A ratio is said to be given, when a ratio of a given magnitude 
to a given magnitude which is the same ratio with it can be 
found. 

III. 
Rectilineal figures are said to be given in species, which have 
each of their angles given, and the ratios of their sides given. 

Points, lines, and spaces, are said to be given in position, which 
have always the same situation, and which are either actually 
exhibited, or can be found. 

A. 
An angle is said to be given in position, which is contained by 
straight lines given in position. 

V. 
A circle is said to be given in magnitude, when a straight line 
from its centre to the circumference is given in magnitude. 

A circle is said to be given in position and magnitude, the 
centre of which is given in position, and a straight line from 
it to the circumference is given in magnitude. 
VII. 
Segments of circles are said to be given in magnitude, when 
the angles in them, and their bases, are given in magnitude. 
VIII. 
Segments of circles are said to be given in position and mag- 
nitude, when the angles in them are given in magnitude, and 
their bases are given both in position and magnitude. 
IX. 
A magnitude is said to be greater than another by a given 
magnitude, when this given magnitude being taken trom it, 
the remainder is equal to the other magnitude. 

Aa4 X. A. 



360 



EUCLID'S 



See N, 



» 1. dcf. 
dat. 



2. 
SeeN. 



• 1. def. 



11. 



X. 

A magnitude is said to be less than another by a given magni- 
tude, when this given magnitude being added to it, the whole 
is equal to the other magnitude. ' 

PROPOSITION I. 



HE ratios of given magnitudes to one another 



IS 



A B C I) 



Let A, B be two given magnitudes, the ratio of A to B is 
given. 

Because A is a given magnitude, there may 
* be found one equal to it •, let this be C : 
And because B is given, one equal to it may 
be found j let it be D j and since A is equal 
to C, and B to.D : therefore'' A is to B, as 
C to D ; and consequently the ratio of A to 
B is given, because the ratio of the given 
magnitudes C, D, which is the same with it, 
has been found. 

PROP. II. / 

If a given magnitude has a given ratio to another 
magnitude, ''and if unto the two magnitudes by 
" which tlie given ratio is exhibited, and the given 
" magnitude, a fourth proportional can be found ;" 
the other magnitude is given. 

Let the given magnitude A have a given ratio to the mag- 
nitude B ; if a fourth proportional can be found to the three 
magnitudes above named, f> is given in magnitude. 

Because A is given, a magnitude may be 
found equal to it^-, let this be C: And be- 
cause the ratio of A to B is given, a ratio 
which is the same with it may be found j let 
this be the ratio of the given magnitude E 
to the given magnitude F : Unto the magni- 
tudes E, F, C, find a fourth proportional 
D, which, by the hypothesis, can be done. 
Wherefore, because A is to B, as E to F j and 
as £ to F4 so is C to D 1 A is'' to B, as C to 

* Tbe figures in the margin shew the number pf the propoikioni ifi Uit other edi- 

tiOMt . 



AB C 



D 
E 



DATA. 361 

D. But A is equal to C ; therefore" B is equal to D. The ' ^*- ^• 
magnitude B is therefore given*, because a magnitude D equal * !• ^^• 
to it has been found. 

The limitation within the inverted commas is not in the 
Greek text, but is now necessarily added ; and the same 
must be understood in all the propositions of the book which 
depend upon this second proposition, where it is not expressly 
mentioned. See the note upon it. 



PROP. III. 3. 

1 F any given magnitudes be added together, their 
sum shall be given. 

Let any given magnitudes AB, BC be added together, their 
sum AC is given. 

Because AB is given, a magnitude equal to it may' be fouad j » 1. d^f. 
let this be DE : And because BC is 

given, one equal to it may be found ; j\. [y Q 

let this be EF : Wherefore, because 

AB is equal to DE, and BC equal 1) jr p 

to EF ; the whole AC is equal to [ " 

the whole DF : AC is therefore given, because DF has been 
found which is equal to it. 



PROP. IV. 4, 

j[ F a given magnitude be taken from a given mag- 
nitude ; the remaining magnitude shall be given. 

From the given magnitude AB, let the given magnitude 
AC be taken ; the remaining magnitude CB is given. 

Because AB is given, a magnitude equal to it may* bc^^j^ 
found ; let this be DE : And be- 
cause AC is given, one equal to it .Y C R 

may be found ; let this be DF : ~ ' ' ' 

Wherefore because AB is equal to ^^ ir i-. 

DE, and AC to DF ; the remain- ii fe E 

der CB is equal to the remainder 

FE. CB is therefore given% because FE which is equal to 

it has been found. 



* 4. dat. 



362 E U C D I D ' S 

12. PROP. V. 

Set N. X F of three magnitudes, the first together with the 
second be given, and also the second together M'ith 
the third ; either the first is equal to the third, or 
one of them is greater than the other by a given 
magnitude. 

Let AB, BC, CD be three magnitudes, of which AB toge- 
ther with BC, that is, AC, is given ; and also BG together 
with CD, that is, BD, is given. Either A B is equal to CD, or 
one of them is greater than the other by a given magnitude. 

Because AC, BD are each of them given, they are either 
equal to one another, or not equal, a U f [\ 

First, let them be equal, and be- ^ ■^ — 

cause AC is equal to BD, take away the common part BC ', 
therefore the remainder AB is equal to the remainder CD. 

But if they be unequal, let AC be greater than BD, and 
make CE equal to BD. Therefore CE is given, because BD 
is given. And the whole AC is 
given i therefore" AE the remain- * ' TT P r r\ 

der is given. And because EC is ^^ " ~ ^^ ii 

« «qual to BD, by taking BC from 

both, the remainder EB is equal to the remainder CD. And 
AE is given ; wherefore AB exceeds EB, that is, CD, by the 
given magnitude AE. 

i- PROP. VI. 

seeN. 1 F a magnitude has a given ratio to a part of it, it 
shall also have a given ratio to the remaining part of it. 

Let the magnitude AB have a given ratio to AC a part of 
it J it has also a given ratio to the remainder BC. 

Because the ratio of AB to AC is given, a ratio may be 
> 2. Hef. found* which is the same to it : Let this be the ratio of DE, 
a given magnitude to the given mag- 
nitude DF. And because DE, DFare A C B 

M. dat. gi^sn> th^ remainder FE is*" given : 

And because AB is to AC, as DE to ]) V £ 

<E. 5, DF, by conversion'^ AB is to BC, as 

DE toEF. Therefore the ratio of AB to BC is given, be- 
cause the ratio of the given magnitudes DE, EF, which is the 
same with it, has been found. 

CoR. 



DATA. 363 

Cor. From this it follows, that the parts AC, CB have a 

fiven ratio to one another : Because as AB to BC, so is DE to 
;F i by divisions AC is to CB, as DF to FE ; and DF, FE " n. 5. 
are given ; therefore* the ratio of AC to CB is given. * '• ^^'f- 

PROP. VII. 6. 

J.F two magnitudes which have a given ratio to ^e* >•'• 
one another, be added together; the whole magni- 
tude shall liave to each of them a given ratio. 

Let the magnitudes AB, BC, which have a given ratio to 
one another, be added together ; the whole AC has to each 
of the magnitudes, AB, BC a given ratio. 

Because the ratio of AB to BC is given, a ratio may be 
found* which is the same with it ; let this be the ratio of the »2. drf. 
given magnitudes DE, LF : And 

because DE, EF are given, the A ~ R C 

whole DF is given'': And because «> 3.dat, 

as AB to BC, so is DE to EF ; by ]) "R F * 

composition- AC is to CB as DF to "^ ^5- •^• 

FE; and, by conversion'^, AC is to AB, as DF to DE '."E.a. 
Wherefore because AC is to each of the magnitudes AB, BC, 
as DF to each of the others DE, EF ; the ratio of AC to 
each of the magnitudes AB, BC is given^. 

PROP. VIII. 7. 

IF a given magnitude be divided into two parts See n. 
which have a given ratio to one another, and if a 
fourth proportional can be found to the sum of the 
two masfnitudes bv which the given ratio is exhi- 
bited, one of tliem, and the given magnitude ; each 
of the parts is given. 

Let the given magnitude AB be divided into the parts AC,. 
CB, which have a given ratio to one another ; if a fourth pro- 
portional can be found to the above- a C 15 
named magnitudes; AC and CB are ' 
each of them given. -p^ ^ 

Because the ratio of AC to CB is ^ A — ±2 

given, the ratio of AB to BC is given * therefore a ratio . 7^ j^^ 

which 



364 



2. def. 



« 2. dat. 
* 4.dat. 



E U C L I D'S 

which is the same with it can be found'', let this be the rati© 
of the given magnitudes, DE, EF: 

And because the given magnitude ^ Q 33 

AB has to BC the given ratio of DE ' "" 

to EF, if unto DE, EF, AB a fourth j) F F 

proportional can be found, this which "^ ~ 

is BC is given=^; and because AB is given, the other part AC 
is given''. ny^' •' . ' 'c 

In the same manner, and with the like limitation, if the dif- 
ference AC of two magnitudes AB, BC, which have a given 
ratio be given i each of the magnitudes AB, BC is given. 



PROP. IX. 



JVl AGNITUDES which have given ratios to the 
same magnitude, have also a given ratio to one 
another. 



Let A, C have each of them a given ratio to B : A has a 
given ratio to C. 

Because the ratio of A to B is given, a ratio which is the 
• '2. def. same to it may be found* ; let this be the ratio of the given 
magnitudes_D, E : And because the ratio of B to C is given, 
a ratio which is the same with it may be found* ; let this be 
the ratio of the given magnitudes 
F, G : To F, G, E find a fourth 
proportional H, if it can be done ; 
and because as A is to B, so is D 
to E; andasBtoC, so is (I'' to G, 
and so is) E to H ; ejc aequali, as 
A to C, so is D to H: Therefore A B C D E H 
the ratio of A to C is given% be- 
cause the ratio of the given magni- F Q. 
tudcs D and H, which is the same 
• with it has been found: But if a 
fourth proportional to F, G, E can- 
not be found, then it can only be said that the ratio of A to C 
is compounded of the ratios of A to B, and B to C, that is, of 
the given ratios of D to E, and F G. 



DATA. 36s 



PROP. X. 9. 

Xf two or more magnitudes have given ratios to 
one another, and if they have given ratios, though 
they \ye not the same, to some other magnitudes : 
these other magnitudes shall also ha\e given ratios 
to one another. 

Let two or more magnitudes A, B, C have given ratios to 
one another ; and let them have given ratios, though they be 
not the same, to some other magnitudes D, E, V : The mag- 
nitudes D, E, F have given ratios to one another. 

Because the ratio of A to B is given, and likewise the ratio 

of A to D i therefore the ra- 

tio of D to B is given*; but -^ ^ » 9. «Ut. 

the ratio of B to £ is given, r» -r 

therefore* the ratio of D to 

E is given: And because the q -p 

ratio of B to C is given, and 

also the ratio of B to E ; the ratio of E to C is given* : And 
the ratio of C to F is given ; wherefore the ratio of E to F is 
given : D, E, F have therefore given ratios to one another. 



PROP. XI. 22. 

JlF two magnitudes have each of them a given 
ratio to anotiber magnitude, both of them together 
shall have a given ratio to that other. 

Let the magnitudes AB, BC have a given ratio to the mag- 
nitude D, AC has a given ratio to the same D. 

Because AB, BC have each of 

them a given ratio to D, the ratio A R C 

of AB to BC is given*: And by •9.dat. 

composition, the ratio of AC to CB J) 

is given^ : But the ratio of BC to » 7. dat. 

D is given ; therefore* the ratio of AC to D is given. 



366; E U € L I D'S 

23. PROP. XII. 

SeeN. If the. whole have to the whole a given ratio, 
and the parts have to the parts given, but not the 
same ratios: Every one of them, whole or part, 
shall have to every one a ^reat ratio. 

Let the whole AB have a given ratio to the whole CD, and 
the parts AE, EB have given, but not the same, ratios to the 
parts CF, FD : Every one shall have to every one, whole or 
part, a given ratio. . 

Because the ratio of AE to CF is given, as AE to CF, so 
make AB to CG ; the ratio therefore of A B to CG is given : 
wherefore the ratio of the remainder EB to the remainder 

« 19. 5. FG is given, because it is the same' with the ratio of AB to 
CG: AndtheratioofEBto FDis ~ 

given, wherefore the ratio of FD *^ ^ Js 

to FG is given** ; and, by conver 

sion, the ratio of FD to DG is C F G D 

given*^ : And because AB has to ~ ' ^ 

each of the magnitudes CD, CG a given ratio, the ratio of 
CD to CG is given**, and therefore <^ the ratio of CD to DG 
is given : But the ratio of GD to DF is given, vi^herefore'' the 

••cor. 6. ratio of CD to DF is given, and consequently"^ the ratio of 
^*'- CF to FD is given ; but the ratio of CF to AE is given, as 

•iO. dat. ^Iso the ratio of FD to EB ; vv^herefore"^ the ratio of AE to 

f ^ ^^j EB is given ; as also the ratio of AB to each of them*^. The 
ratio therefore of every one to every one is given. 

24. PROP. XIII. 

See N. J^ p ^l^g ^jj.j.|. ^^ three proportional straight lines has 
a given ratio to the third, the first shall also have a 
given ratio to the second. 

Let A, BjC be three proportional straight lines, that is, as 
A to B, so is B to C ; if A has to C a given ratio, A shall also 
have to B a given ratio. 

Because the ratio of A to C is given, a ratio which is the 
♦ 9. def. same with it may be found^ let this be the ratio of the given 
^13, s. straight lines D, Ej and between D and E find a** mean 

proportional 



«> 9. dit. 
•^ 6. dat. 



DATA. 

proportional F j therefore the rectangle contained by D and 

K is equal to the square of F, and the reft- 

angle D, E is given, because its sides D, E 

are given ; wherefore the square of F, and 

the straight line F is given : And because as 

A is to C, so is D to E } but as A to C, so 

is<^ the square of A to the square of B ; and 

as D to E, so is*= the square of D to the ^ J^ C *2 

square of F : Therefore the square'^ of A is to 

the square of B, as the square of D to the 

square of F: As therefore"" the straight line 

A to the straight line B, so is the straight line 

D to the straight line F; therefore the ratio 

of A to B is given*, because the ratio of the 

given straight lines D, F, which is the same 

with it, has been found. 



367 



D F K 



2. cor. 

'JO. 6. 

MI. 5. 

« 22. 6. 



9. daf. 



PROP. XIV A. 

XF a magnitude, together with a given magnitude, sceN. 
has a given ratio to another magnitude ; the excess 
of this other ma;>:nitude ahove a s^iven maonitude 
has a o;n-en ratio to the first masrnitude : And iC 
the excess of a mao-nitude above a srivcn ma«:nitude 
has a given ratio to another magnitude ; this other 
magnitude, together with a given magnitufle, has 
a given ratio to the first magnitude. 

Let the magnitude AB, together with the given magnitude 
BE, that is, AE, have a given ratio to the magnitude CD : the 
excess of CD above a given magnitude has a given ratio to AB. 

Because the ratio of AE to CD is given, as AE to CD, so 
mate BE to FD; therefore the ratio of BE to FD is given, 
and BE is given ; wherefore FD 

is given^' : And because as AE »0 3 ^ »«.<Jat, 

to CD, so is B¥. t® FD, the re- 
mainder A B is ''to the remainder C F D '19.5, 

'CF, as AE to CD : But the ratio 

of AE to CD is given; therefore the ratio of AB to CF is 
given ; that is, CF the excess of CD above the given magni- 
tude FD has a given ratio to AB. 

Next, Let the excess of the magnitude AB above the given 
magnitude BE, that is, let AE havea given ratio to the mag- 
nitude, 



368 EUCLID'S 

nitude CD ; CD together with a given magniiud^ has a 
given ratio to AB. 

Because the ratio of AE to CD is given, as AE to CD, so 
makeBEtoFDjthereforetheratioof -r? -n 
BE to FD is given, and BE is given ^ J^i S 

•2.dat. wherefore FD is given": And because 

as AE to CD, so is BE to FD, AB is -r, n T 

' ^^- ^' to CF, as-^ AE to CD : But the ratio ^ "M— ^ 

of AE to CD is given, therefore the ratio of A B to CF is 
given : that is, CF which is equal to CD, together with the 
given magnitude DF, has a given ratio to AB. 

B. PROP. XV. 

SeeN. J p r^ magnitude, together with that to which ano- 
ther magnitude has a given ratio, be given ; the 
sum of this other, and that to which the first 
magnitude has a given ratio, is given. 

Let AB, CD be two magnitudes of 'which AB together with 
BE to which CD has a given ratio, iji given j CD is given to- 
gether with that magnitude to which AB has a given ratio. 
Because the ratio of CU to BE is given, as BF to CD, so 
make AE to FD ; therefore the ratio of AE to FD is given, 

» 2.dat. and AE is given, wherefore* FD 

is given : And because as BE to A Jj K 

•'Cor.i9.5cD,so isAE toFD: AB is" to 

FC, as BE to CD: And the ratio y CT> 

of BE to CD is piven, wherefore ' "~~ 

the ratio of AB to FC is given : And FD is given, that is, CD 
together with FC to which AB has a given ratio is given. 

10. PROP. XVI. 

sceN. J[p t]-,g excess of a' magnitude above a given mag- 
nitude has a given ratio to another magnitude ; the 
excess of both together above a given magnitude 
shall have to that other a given ratio : And if the 
excess of two magnitudes together above a given 
magnitude, has to one of tliem a given ratio ; either 
the excess of the other above a given magnitude 
has to that one a given ratio ; or tiie other is given 
together with the magnitude to wliich tliat one has 



a given ratio. 



/Ct 



DATA. 369 

Let the ejtcess of the magnitude AB above a given magni- 
tude, have a given ratio to the magnitude BC ; the excess of 
AC, both of them together, above the given niagnitude, has a 
given ratio to BC. / 

Let AD be the given magnitude, the excess of AB above 
which, viz. DB, has a given ratio 

to BC : And beciuse'DB has a A. T) B C 

given ratio to BC, the ratio of 

DC to CB is given% and AD is given j therefore DC, the ex- » 7. dat. 
cess of AC above the given magnitude AD, has a given ratio 
to BC. 

Next, Let the excess of two magnitudes AB, BC together, 
above a given magnitude, have to 

one of them BC a given ratio j ei- A D B !E C 

ther the excess of the other of them 

AB above the given magnitude shall have to BC a given ra- 
tio J or AB is given, together with the magnitude to which 
£C has a given ratio. 

Let AD be the given magnitude, and first let it be less 
than AB ; and because DC the excess of AC above AD has 
a given ratio to BC, DB has'' a given ratio to BC ; that is, » «or. 6. 
DB the excess of AB above the given magnitude AD has a ^^- 
given ratio to BC. 

But let the given magnitude be greater than AB, and make 
AE equal to it ; and because EC, the excess of AC above AE 
has to BC a given ratio, BC has' a given ratio to BE ; and ' 6. dat. 
because AE Is given, AB together with BE to which BC has 
a given ratio is given. 

PROP. XVIL H. 

J.F the excess of a magni' ade above a given mag- SeeN. 
iiitude has a given ratio to another magnitude ; the 
excess of the same first magnitude above a given 
magnitude, shall have a given ratio to both the 
magnitudes together. And if the excess of either 
of two magnitudes above a given magnitude has a 
given ratio to both magnitudes together ; the ex- 
cess of the same above a given magnitude shall 
I have a given ratio to the other. 

Let the excess of the magnitude AB above a given magni- 
tude have a given ratio to the magnitude BC ; the excess of 
AB above a given magnitude has a given ratio to AC. 

B b Let 



370 EUCLID'S 

Let AD be the given magnitutle ; and because DB, the ex- 
cess of AB above AD, has a given ratio to BC ; the ratio of 

* 7- dat. DC to DB is given* : Make the ratio of AD to DE the same 
with this ratio ; therefore the ra- . T T\ T> n 
tio of AD to DE is given ; and ^ - V Y ^ ^ ^ 

•> 2. dat. AD is given, wherefore'' DE and the remainder AE are given: 

' 12. 5. And because as DC to DB, so is AD to DE, AC is-^ to EB, 
as DC to DB i and the ratio of DC to DB is given j where- 
fore the ratio of AC to EB is given : And- because the ratio 
of EB to AC is given, and that AE is given, therefore EB 
the excess^of AB above the given magnitude AE, has a given 
ratiato AC. 

Next, Let the excess of AB above a given magnitude have a ■ 
given ratio to AB and BC together, that is, to AC ; the ex- 
cess of AB above a given magnitude has a given ratio to BC. 
Let AE be the given ma:gnitude j and because EB the 
excess of AB above AE has to AC a given ratio, as AC to 
EB so make AD to DE ; therefore the ratio of AD to DE is 

d g_ dat. given, as also<^ the ratio of AD to AE : And AE is given, 
wherefore^ AD is given : And because, as the whole AC, to 

e 19 5^ the whole EB, so is AD to DE, the remainder DC is*" to the 
remainder DB, as AC to EB ; and the ratio of AC to EB is 

f Cor. 6. given ; wherefore the ratio of DC to DB is given, as also*^ the 
tiat, ratio of DB to BC : And AD is given ; therefore DB, the 
excess of AB above a given magnitude AD, has a given ratio 
toBC. 



u. PROP. xvin. 



1 



» i . dat. 



F to each of two raagnitudes, which have a given 
ratio to one anotlier, a given magnitude be added ; 
the wholes shall either have a given ratio to one 
another, or the excess of one of them above a given 
magnitude shall have a given ratio to the other. 

Let the two magnitudes AB, CD ha,ve a Sfiven ratio to one 
another, and to AB let the given magnitude BE be added, 
and the given magnitude DF to CD; The wholes AE, CF 
either have a given ratio to one another, or the excess ot one of 
them above a given magnitude has a given ratio to the other*. 

Because BE, DF are each of them given, their ratio is given, 

and 



DATA. 371 

and if this ratio be the same with the A "R V 

ratio g; AB to CD, the ratio of AE '^ "^ ~ 

to CF, which is the samek with the #> 7^ -tr* " 12. 5. 

given ratio o( AB to CD, shall be ^^-^ £- 

given. 

But if the ratio of BE to DF be not the same with the ratio 
of AB to CD, either it is greater than the ratio of AB to 
CD, or, by inversion, the ratio of DF to BE is greater than 
the ratio of CD to AB : First, let 

the ratio of BE to DF be greater A !B C- TT^ 

than the ratio of AB to CD j and 

as AB to CD, so make BG to DF ; Q J) 

therefore the ratio of BG to DF is 

given; and DF is given, therefore"^ BG is given: And '2.dat, 
because BE has a greater rati > to DF than ( AB to CD, that is, 
than) BG to DF, be is greater"^ trian BG : And because as * '0-5. 
AB to CD, so is BG to DF ; therefore AG is' to CF, as AB 
to CD : But the ratio of AB to CD is giv°n, .vhercfore the 
ratio of AG to CF is given ; and because Bti^ BG are each of 
them given, GE is given : Therefore AC', the excess of AE 
above a given magnitude GE, has a given ratio to CF. The 
other case is demonstrated in the same manner, 



PROP. XIX. 15. 



JlF from each of two magnitudes, which have a given 
latio to one another, a ariven maarnitude be taken, 
the remainders shall either have a given ratio to one 
another, or tlie excess of one of them above a given 
magnitude, shall have a given ratio to the other. 

Let the maixnitudes AB, CD have a criven ratio to one an- 
other, and from AB let the given magnitude AE be taken, 
and from CD the given magnitudes CF ; The remainders EB, 
FD shall either have a given ratio to one another, or the 
excess of one of them above a 

given magnitude shall have a ^ iji S 

given ratio to the other. 

Because AE, CF are each of C Y T> 

them given, their ratio is - - 

_'iven*i and if this ratio be the same with the ratio of AB to » 1 ,1,^, 

Bb2 CD, ■ 



37* EUCLID'S 

CD, the ratio of the remainder £B to the remainder FD, 
» 19. 5. which is the same** with the given ratio of AB to CD, shall 
be given. 

But if the ratio of AB to CD be not the same with the 
ratio of AE to CF, either it is greater than the ratio of AE 
to CF, or, by inversion, the ratio ©f CD to AB is greater than 
the ratio of CF to AE : First, let the ratio of AB to CD be 
greater than the ratio of AE to CF, and as AB to CD, so 
make AG to CF : therefore the 

ratio of AG to CF is given, and A, E G ^ 

' 2. dat. Q£ is given, wherefore' AG is 

given : And because the ratio of C y D 

AB to CD, that is, the ratio of 

AG to CF, is greater than the ratio of AE to CF ; AG is 
• l», 5. greater** than AE ; and AG, AE are given, therefore the 
remainder EG is given ; and as AB to CD, so is AG to CF, 
and so is** the remainder GB to the remainder FD j and the 
ratio of AB to CD is given : W herefore the ratio of GB to 
FD is given ; therefore GB, the excess of EB above a given 
magnitude EG, has a given ratio to FD. In the same man> 
ner the other case is demonstrated. 

16. PROP. XX. 

XF to one of two magnitudes which have a given 
ratio to one another, a given magnitude be added, 
and from the other a given magnitude be taken ; 
the excess of the sum above a given magnitude 
shall have a given ratio to the remainder. 

Let the two magnitudes AB, CD have a given ratio to one 
another, and to AB let the given magnitude E A be added, and 
from CD let the given magnitude CF be taken j the excess of 
the sum EB above a given magnitude has a given ratio'to the 
remainder FD. 

Because the ratio of AB to CD is given, make as AB to CD, 
so AG to CF : Therefore the ratio of AG to CF is given, and 
CF is given, wherefore* AG is 

given: and EA is given, there- K A G P> 

fore the whole EG is given: And 
^ because as AB to CD, so is AG q y jy 

' to CF, and so is"* the remainder ' 

GB to the remainder FD ; the ratio of GB to FD is given. 
And EG is given, therefore GB, the excess of the sum EB 

I abo\ 



DATA. 373 

above the given magnitude EG, has a given ratio to the 
remainder FD. 

PROP. XXI. c. 

XF two magnitudes have a given ratio to one ano- SeeN. 
ther, if a given magnitude be added to one of them, 
and the other be taken from a given magnitude ; 
the sum, together with the magnitude to which the 
remainder has a given ratio, is given : And the 
remainder is given together with the magnitude to 
which the sum has a sciven ratio. 



o* 



Let the two magnitudes AB, CD have a given ratio to one 
another ; and to AB let the given magnitude BE be added, and 
let CD be taken from the given magnitude FD : The sum AE 
is given together with the magnitude to which the remainder 
FC has a given ratio. 

Because the ratio of AB to CD is given, make as AB to 
CD, so GB to FD : Therefore the ratio of GB to FD is given, 
and FD is given, wherefore GB 

is given*; and BE is given, the G A R E • 3.dat. 

whole GE is therefore given, and 

because as AB to CD, so is GB F C D 

to FD, and so is" GA to FC ; the ' - 1?- 5. 

ratio of GA to FC is given : And AE together with GA is 
given, because GE is given ; therefore the sum AE, together 
with GA, to which the remainder FC has a given ratio, is 
given. The second part is manifest from Prop. 15. 

PROP. XXII. D. 

If two magnitudes have a given ratio to one ano- see n. 
ther, if from one of them a given magnitude be 
taken, and the other be taken from a given mag- 
nitude ; each of the remainders is given together 
with the magnitude to which the other remainder 
has a given ratio. 



Let the two magnitudes AB, CD have a given ratio to one 
another, and from AB let the given magnitude AE be taken, 

B b 3 and 



374 E U C L I D'S 

and let CD be taken from the given magnitude CF : The re- 
nlainder EB is given together vvith the magnitude to which 
the other remainder DF has a given ratio. 

Because the ratio of AB to CD is -given, make as AB to 
CD, so AG to CF: The ratio of AG to CF is therefore given, 

» 2. dat. and CF is given, vi^herefore* AG 

is given ; and AE is given, and .A ^ ]B Q: 

therefore the remainder EG is 

given : And because as AB to Q J) ]p 

b 19. 5. CD, so is AG to CF : And so is" '^ 

the remainder BG to the remainder DF ; the ratio of BGto 
DF is given : And EB together vi'ith BG is given, because 
EG is given: Therefore the remainder EB, together with BG, 
to which DF the other remainder has a given ratio, is given. 
The second part is plain from this and Prop. 15, ' 



20, 

See N. 



PROP. XXIII. 



If from two given magnitudes there be taken mag- 
nitudes which have a given ratio to one another, 
the remainders shall either have a given ratio to one 
anotlier, or the excess of one of them above a given 
magTiitude shall have a given ratio to the- other.' 

Let AB, CD be two given magnitudes, and from them let 
the magnitudes AE, CF, which have a given ratio to one 
another, be taken ; the remainders EB, FD either have a given 
ratio to one another, or the excess of one of them above a 
given magnitude has a given ratio to the other. 
Because AB, CD are each of 

them given, the ratio of AB to A , Iji jjt 

CD is given : And if this ratio 

be the same with the ratio of AE Q T Q 

to CF, then the remainder EB ' 

» 19. 5. has' the same given ratio to the remainder FD. 

But if the ratio of AB to CD be not the same with the ratio 
of AE to CF, it is either greater than it, or, by inversion, 
the ratio of CD to AB is greater than the ratio of CF to AE : 
First, let the ratio of AB to CD be greater than the ratio of 
AE to CP" ; and as AE to CF, so make AG to CD : there- 
fore the ratio of AG to CD is given, because the ratio of 
•> 2. dat. AE to CF is given j and CD is given, wherefore'* AG is 

given i' 



DATA. 

given ; and because the ratio of AB to CD is greater than the 
ratio of (AE to CF, that is. 



37S 



A 



E 



GB 



« 10, 5. • 



C 



D 



than the ratio of) AG to CD; 
AB is greater"^ than AG: 
And AB, AG are given; 
therefore the remainder BG ' 

is given : And because as AE to CF, so is AG to CD, and 
so is* EG to FD ; the ratio of EG to FD is given : And GB ' '*• ^• 
is given ; therefore EG, the excess of EB above a given mag- 
nitude GB, has a given ratio to FD. The other case is 
shewn in the same way. 



PROP. XXIV. 



J 3, 



1 F there be three magnitudes, the first of which has 
a given ratio to the second, and the excess of the 
second above a given magnitude has a given ratio to 
tlie third ; the excess of the first above a given mag- 
nitude shall also have a given ratio to the third. 

Let AB, CD, £, be the three magnitudes of which AB has 
a given ratio to CD ; and the excess of CD above a given 
magnitude has a given ratio to E : The excess of AB above a 
given magnitude has a given ratio to E. 

Let CF be the given magnitude, the excess of CD above 
which, viz. FD, has a given ratio to K : And because the ratio 
of AB to CD is given, as AB to CD, so make 
AG to CF ; therefore the ratio of AG to CF ' 
is given : And CF is given, wherefore* AG is 
given : And because as A B to CD, so is AG p|_ C 
to CF, and so is" GB to FD ; the ratio of GB ^^ 
to FD is given. And the ratio of FD to E is 
given, wherefore "^ the ratio of GB to E is 
given, and AG is given ; therefore GB the 
excess of AB above a given magnitude AG ]g J) 
has a given ratio to E. 

CoR. I. And if the first has a given ratio to the second, 
and the excess of the first above a given magnitude has a 
given ratio to the third ; the excess of the second above a given 
magnitude shall have a given ratio to the third. For, if the 
second be called the first, and the first the second, this corol- 
lary will be the same with the proposition. 

Bb4 CoR. 



SeeN. 



M9. 5. 



^ 9. dat. 



376 EUCLID'S 

Cor. 2. Also, if the first has a given ratio to the second, 
and the excess of a third above a given magnitude has also 
a given ratio to the second, the same excess shall have a given 
ratio to the first j as is evident from the 9th dat. 

17. PROP. XXV. 

I F there be three magnitudes, the excess of the first 
whereof above a given magnitude has a given ratio 
to the second j and the excess of a third above a 
given magnitude has a given ratio to the same se- 
cond ; The first shall either have a given ratio to the 
third, or the excess of one of them above a given 
magnitude shall have a given ratio to the other. 

Let AB, C, DE be three magnitudes, and let the excesses 
of each of the two AB, DE above given magnitudes have 
given ratios to C j AB, DE either have a given ratio to one 
another, or the excess of one of them above a given magni- 
tude has a given ratio to the other. 

Let FB the excess of AB above the given magnitude AF 
have a given ratio to C ; and let GE the ex- 
cess of DE above the given magnitude DG a 
have a given ratio to C ; and because FB,GE 
have each of them a given ratio to C, they jp 

•§, dat. have a given ratio* to one another. ButtoFB, 
GE the given magnitudes AF, DG are 

» 18. dat. added ; therefore'' the whole magnitudes AB, 
DE have either a given ratio to one another, 
or the excess of one of them above a given B 
magnitude has a given ratio to the other. 

18- PROP, XXVL 



C 



D 
G 



_l F there be three magnitudes, tlie excesses of one of 
which above given magnitudes have given ratios 
to the other two magnitudes ; these two shall either 
have a given ratio to one another, or the excess of 
one of them above a given magnitude shall have a 
given ratio to the other, 

Let 



DATA. 



377 



A 


C 

a 

K 


L 
• E 


B 


D 


F 



Let AB, CD, EF be three magnitudes, and let GD the 
excess of one of them CD above the given magnitude CG 
have a given ratio to AB ; and also let KD the excess of the 
same CD above the given magnitude CK have a given ratio to 
EF, either AB has a given ratio to EF, or the excess of one of 
them above a given magnitude has a given ratio to the other. 

Because GD has a given ratio to AB, as GD to AB, so 
make CG to HA ; therefore the ratio of CG to HA is given ; 
and CG is given, wherefore* HA is given : And because as • 2. dat. 
GD to AB, so is CG to HA, and so is" CD to HB ; the ratio k it. s. 
of CD to HB is given: Also because KD has a given ratio 
to EF, as KD to EF, so make CK to LE i ttj 
therefore the ratio of CK to LE is given ; and 
CK is given, wherefore LE* is given : And 
because as KD to EF, so is CK to LE, and 
so" is CD to LF ; the ratio of CD to LF is 
given : But the ratio of CD to HB is given, 
wherefore*^ the ratio of HB to LF is given : ""^ *^ '9. dat. 

and from HB, LF the given magnitudes HA 
LE being taken, the remainders aB, EF shall 
either have a given ratio to one another, or the excess of one 
of them above a given magnitude has a given ratio to the other**. * I9. dat. 

Another Demonstration. 

Let AB, C, DE be three magnitudes, and let the excesses 
of one of them C above given magnitudes have given ratios 
to AB and DE ; either AB, DE have a given ratio to one 
another, or the excess of one of them above a given magnitude 
has a given ratio to the other. 

Because the excess of C above a given magnitude Jias a 
given ratio to AB ; therefore * AB together with a given mag-* i*. *t. 
nitude has a given ratio to C : Let this given ip 
magnitude be AF, wherefore FB has a given 
ratio toC : Also because the excess of C above 
a given magnitude has a given ratio to DE ; 
therefore * DE together with a given magni- 
tude has a given ratio to C : Let this given 
magnitude be DG, wherefore GE has a given 
ratio to C : And FB has a given ratio to C, therefore" the ratio ' 9. «lat, 
of FB to GE is given : And from FB, GE the given magni- 
tudes AF, DG being taken, the remainders AB, DE either 
have a given ratio to one another, or the excess of one of them 
above a given magnitude has a given ratio to the other*". ' 19. dat. 



B 



G 
D 

Cl E 



378 



E U C L I D'S 



1?. 



- PROP. XXVII. 

-1 F there be three magnitudes, the excess of tlie 
first of which above a given magnitude has a given 
ratio to the secopd ; and the excess of the second 
above a given magnitude has also a given ratio to 
the third : The excess of the first above a given 
maw-nitude shall have a eiven ratio to the third. 



» 2. dat. 



19.3. 



9. dat. 



Let AB, CD, E be three magnitudes, the excess of the first 
of which AB above the given magnitude AG, viz. GB, has a 
given ratio to CD ; and FD the excess of CD above the given 
magnitude CF, has a given ratio to E : the excess of AB 
above a given magnitude has a given ratio to E. 

Because the ratio of GB to CD is given, as GB to CD, so 
make GH to CF ; therefore the ratio of GH 
to (^F is given ; and CF is given, wherefore=^ 
GH is given ; and AG is given, wherefore 
the whole AH is given : And because as GB 
to CD, so is GH to CF, and so is ^ the re- 
mainder HB to the remainder FD ; the ratio 
of HB to FD is given : And the ratio of FD 
to E is given, wherefore "^ the ratio of HB to 
E is given : And AH is given j therefore HB 
the excess of AB above a given magnitude AH ^as a giveii 
ratio to E. 



A 






G 


C 




H 


- F 


- 


B 


D 


E 



" 24. dut. 



*' Otherwise, 

Let AB, C, D be three magnitudes, the excess EB of the 
first of which AB above the given magnitude AE has a given 
ratio to C, and the excess of C above a given 
magnitude has a given ratio to D ; The ex- ^ 
cess of AB above a given magnitude has a 
given ratio to D. E 

Because EB has a given ratio toC, and the 
excess ofC above a given magnitude has a y, 
given ratio to D; therefore'' the excess of EB 
above a given magnitude has a given ratio to 
D : Let this given magnitude be EF; therefore g 
FB the excess of EB above EF has a given 
ratio toD: And AF is given, because AE, EF 



c 



D 



are 



D A T A. 379 

are given : Therefore FB the excess of AB above a given 
magnitude AF has a given ratio to D." 



PROP. XXVIII. ^ 25. 



I 




F two lines given in position cut one another, the See n. 
point or points in which they cut one another are 
given. 

Let two lines AB, CD, given in position, cut one another 
in the point E j the point E is 
given. 

Because the lines AB, CD 
are given in position, they have 
always the same situation*, and 
therefore the point, or points, in 
which they cut one another have 
always the same situation : And be- 
cause the lines AB, CD can be 
found % the point, or points, in 
which they cut one another, are 
liicewise found ; and therefore are 
given in position*. 

PROP. XXIX. 2d. 

IF the extremities of a straight line be given In 
position ; the straight line is given in position and 
magnitude. 

Because the extremities of the straight line are given, they 
can be found * : Let these be the points, A, B, between which a 4 ^^f^ 
a straight line AB can be drawn ''i "i.Posttt- 

this has an invariable position, because J^ J^ late. 

between two given points there can 

be drawn but one straight line : And when the straight line 
AB is drawn, its magnitude is at the same time exhibited, or 
given : Therefore the straight line AB is given in position 
and. magnitude. 



38o' EUCLID'S 



2'^. ' PROP. XXX. 

IF one of the extremities of a straight line given 
in position and magnitude be given ; the other ex- 
tremity shall also be given. 

Let the point A be given, to wit, one of the extremities of a 

straight line given in magnitude, and which lies in thd straight 

line AC given in position ; the other extremity is also given. 

Because the straight line is given in magnitude, one equal 

• 1. def. to it can be found"; let this be the straight line D : From the 
greater straight line AC cut off AB 

equal to the lesser D : Therefore the A, B C 

other extremity B of the straight line 

AB is found : And the point B has jy 

al ways the same situation ; because any —— — — — 

other point in AC, upon the same side of A, cuts off between 

it and the point A a greater or less straight line than AB, that 

'*• def. is, than D: Therefore the point B is given'': And it is plain 
another such point can be found in AC, produced upon the 
other side of the' point A. 



'*• PROP. XXXL 

J-F a straight line be drawn through a given point 
parallel to a straight line given in position ; that 
straight line is given in position. 

Let A be a given point, and BC a straight line given in 
position J the straight line drawn through A parallel to BC is 
given in position. 

• 31. 1. Through A draw* the straight line 

DAE parallel to BC; the straight D A E 

line DAE has always the same posi- 
tion, because no other straight line ]^ (^ 
can be drawn through A parallel to 
BC : Therefore the straight line DAE which has beep found 

• 4 def. is given^ in position. 




DATA. 381 

PROP. XXXII. 29. 

X F a straight line be drawii to a given point in a 
straight line given in position, and makes a given 
angle with it ; that straight line is given in position. 

Let AB be a straight line given in position, and C a given 
point in it, the straight line drawn 
to C, which makes a given angle 
with CB, is given in position. 

Because the angle is given, one 
equal to it can be found* j let 

this be the angle at D, at the given 

point C, in the given straight A. 
line AB, make '' the angle ECB 
equal to the angle at D : There- 
fore the straight line EC has al- 
ways the same situation, because 
any other straight line FC, drawn 

to the point C, makes with CB a greater or less angle than 
the angle ECB, or the angle at D : Therefore the straight 
line EC, which has been found, is given in position. 

It is to be observed, that there are two straight lines EC, 
GC upon one side of AB that make equal angles with it, and 
which make equal angles with it when produced to the other 
side. 

PROP. XXXIII. 30, 

JLF a straight hne be dra^vn from a given point to a 
straight line given in position, and makes a given 
angle with it, that straight line is given in position. 

From the given point A let the straight line AD be drawn 
to the straight line BC given in position, and make with it a 
given angle ADC : AD is given in _ 
position. IL =A, F 

Through the point A draw" the \ 'Sj.t, 

straightlineEAF parallel to BC J and \ 

because through the given point A the ^ t\ tt- 

straightlineEAFisdrawn parallel to 

BC, which is given in position, EAF is therefore given in 
position'' : And because the straight line AD meets the parallels t. 31. da(. 

BC, 



382 EUCLID'S. 

'29. 1. BC, EF, the angle is EAD equal<= to the angle ADC j anH 
ADC is given, wherefore also the angle EAD is given : 
Therefore, because the stjaight line DA is drawn to the given 
point A in the straight line EF given in position, and makes 
with It a given angle EAD, AD is given^ in position. 



" 32. dat. 



SceN. 



PROP. XXXIV. 

XF from a given point to a straight line given in 
po.sition, a straight line be drawn which is given in 
magnitude ; the same is also given in position. 

Let A be a given point, and BC a straight line given in 
position, a straight line given in magnitude, drawn from the 
point A toBC is given in position. 

Because the straight line is given in magnitude, one equal to 

»i. def. it can be found'^j let this be the straight line D : From the 
point A draw AE perpendicular to BC : J\^ 

and because AE is the shortest of all the 
straightlines which can be drawn from the 

point A to BC, the straight line D, to ^L 

which one equal is to be drawn from the B ll, C 

point A to BC, cannot be less than AE. J) 

If therefore D be equal to AE, AE is the straight line given 
in magnitude drawn from the given point. A to BC : And it 

*33. dat. is evident that AE is given in position'', because it is drawn 
from the given point A to BC, which is given in position, and 
makes with BC the given angle AEC. 

But if the straight line D be not equal to AE, it must be 
greater than it : Produce AE, and make AF equal to D ; and 
from the centre A, at the distance AF, describe the circle 
GFH, and join AG, AH : because the circle GFH is given 

■^ 6 def. in position^ and the straight line BC is also given in position j 
therefore their intersedion G 

<*28. dat. is given''; and the point A is 
given J wherefore AG is given in 

'29.dat. positions that is, the straight 
line AG given in magnitude, 
(for it is equal to D) and drawn 
from the given point A to the straight line BC given in posi- 
tion, is also given in position : And in like manner AH is 
^ given in position : Therefore in this case there are two straight 

lines 




I 




DATA. 383 

lines AG, AH of the same given magnitude which can be draw" 
from a given peint A to a straight line BC given in position* 

PROP. XXXV. 52. 

If a straight line be drawn between two parallel 
straight lines given in position, and makes given 
jnojles with them, the straisrht line is siven in mas:- 
mtude. 

Let the straight line EF be drawn between the parallels 
AB, CD, which are given in position, and malce the given an- 
gles BEF, EFD : EF is given in magnitude. 

In CD take the given point G, and through G draw^ GH »3i. i. 
parallel to EF : And because CD meets the parallels GH, EF, 
the angle EFD is equal'' to the angle "2?, i. 

HGD : And EFD is a given angle ; A E H R 

wherefore the angle HGD is given : 
and because HG is drawn to the given 

point G, in the straight line CD, given 

in position, and makes a given angle C F G 13 

HGD; the straight line HG is given 
in position'^ : And AB is given in position: therefore the* 32: dat. 
point H is given"^ ; and the point G is also given, wherefore « 28. dat. 
GH is given in magnitude^ : And EF is equal to it, therefore e 09 ,ij,t. 
EF is given in magnitude. 

PROP. XXXVI. 05. 

xF a straight hne given in magnitude be drawn SeeN. 
between two parallel straight lines given in posi- 
tion, it shall make given angles with the parallels. 

Let the straight line EF given in magnitude be drawn be- 
tween the parallel straight lines AB, CD, 
which are given in position : The angles \ JT j-^^ J> 
AEF, EFC^shall be given. ''~ 

Because EF is given in magnitude, a 
straight line equal to it can be found^: • »i.drf 

Let this be G: In AB take a given point p P~K. — T) 

H, and from it draw^ HK perpendicu- t.if 

lar to CD : Therefore the straight line G, ^ 

that 



A E H R 


/\ / 


^ 


C F^ ^OM^ 


t^ND 



384 EUCLID'S 

that is, EF cannot be less than HK : And if G be equal to 
HK, EF also is equal to it ; wherefore EF is at right angles 
to CD J for if it be not, EF would be greater thai HK, which 
is absurd. Therefore the angle EFD is a right, and conse- 
quently a given angle. 

But if the straight line G be not equal to HK, it must be 
greater than it : Produce HK, and take HL, equal to G ; and 
from the centre H, at the distance HL, describe the circle 

• 6- "Jef- MLN, and join HM, HN : And because the circle' MLN, 

and the straight line CD, are given in position, the points M, 
< 28. dat. N are '^ given : And the 

point H is given, wherefore 

the straight lines HM, HN, 
•29.dat. are given in position*': And 

CD is given in position ; 

therefore the angles HMN, 

HNM, are given in posi- ^ 

'A.def. tion^: Of the straight lines 

HM, HN, let HN be that which is not parallel to EF, for EF 

cannot be parallel to both of them ; and draw EO parallel to 

• 34. 1. HN : EO therefore is equale to HN, that is, to G i and EF 

is equal to G j wherefore EO is equal to EF, and the angle 
" 29. 1 , EFO to the angle EOF, that is**, to the given angle HNM, 

and because the angle HNM, which is equal to the angle EFO, 

or EFD, has been found ; therefore the angle EFD, that is, 
" 1. def. the angle AEF, is given in magnitude'' : and consequently 

the angle EFC. 

E. PROP. XXXVH. 

See N. J^ f a Straight line given in magnitude be drawn 
from a point to a straight line given in position, in 
a given angle; the straight line drawn through 
that point parallel to the straight line given in po- 
sition, is given in position. 

Let the straight line AD given in magnitude be drawn from 
the point A to the straight line BC given 
in position, in the given angle ADC : the E A H ^ 
straight line EAF drawn through A pa- 
rallel to BC is given in position. 

In BC take a given point G, and draw GH . 

parallel to AD : And because HG is drawn B DO C 
to a given point G in the straight line BC 

give 



DATA. 



385 



given in position, in a given angle HGC, for it is equal* to* 29. 1. 
the given angle ADC ; HG is given in position'' : But it is " *-. dat. 
given also in magnitude, because it is equal to' AD, which is ^34. 1. 
given in magnitude ; therefore because G, one ofthe extremi- 
ties of the straight line GH given in position and magnitude 
.is given, the other extremity H is given"^ j and the straight <« 30.dat. 
line E AF, which is drawn through the given point H parallel 
to BC given in position, is therefore given^ in position. «3i. da. 



PROP. XXXVIII. 



3-i. 



XF a straight line be drawn from a given point 
to two parallel straight hnes given in position, the 
ratio ofthe segments between the given point and 
the parallels shall be given. 

Let the straight line EFG be drawn from the given point 
E to the parallels AB, CD, the ratio of EF to EG is given. 

From the point E draw EHK perpendicular to CD j and 
because from a given point E the straight line EK is drawn to 
CD which is given in position, in a given angle EKC j EK is 




A 



FH B 



E' 



C 



KGn> 



C G K 



given in position' ; and AB, CD are given in position ; there- »33. At. 
fore^ the points H, K are given : And the point E is given j «> 23. dat, 
wherefore' EH,EK.are given in magnitude, and the ratio'^ofcgg dat 
them is therefore given. But as EH to EK, so is EF to EG, * 1. dat.' 
because AB, CD are parallels j therefore the ratio of EF to 
EG is given. 



PROP. XXXIX. 



35,30. 



F the ratio of the segments of a straight line be- s<?eN-. 
tween a given point in it and two parallel straight 
lines, be given, if one of the parallels be given irr 
position, the other is also given in position. 

C c Froai 



386 E U CL I D'S 

From the given point A, let the straight line AEDbe drawn 
to the two parallel straight lines FG, SC, and let the ratio of* 
the segments AE, AD be given ; if one of the parallels BC be 
given in position, the other FG is also given in position. 

From the point A draw AH perpendicular to BC, and let 
it meet FG in K ; and because AH is drawn from the given 
point A to the straight line BC given in position-, and makes a 




B H TJc: 



B }) II C 



» 33. dat. 

" 28. dat. 
<: 29. dat. 

6 3. dat. 
e 30. dat. 

3I.dat. 




given angle AHD ; AH is given=^ in 

•position ; and BC is lilcewise given in 

positii)n, therefore the point H is 

given'' : The point A is also given ; 

wherefore AH is given in magni- 

tude% and, because FG, BC are 

parallels, as AE to AD, so is AK to 

AH; and the ratio of AE to AD 

is given, wherefore the ratio of AK to AH is given ; but AH 

is given in magnitude, therefore* AK is given in magnitude; 

and it is also given in position, and the point A is given ; 

wherefore'' the point K is given. And because the straight 

line FG is drawn through the given point K parallel to BC 

which is given in position^ therefore'" FG is given in position. 



F~ir 



K (r 



Seen. 



57,58. PROP. XL. 

XF the ratio of the segments of a straio-ht litie into 
which it is cut hy three parallel straight lines, he 
given ; if two of the parallels are given in position, 
the third is also given in position. 

-Let AB, CD, HKbe thf^e|>arallel straight lines, of which 
AB, CD are given in position ; and let the ratio of the seg;- 

meiUs 



DATA. 387 

ments GE, GF into which the straight line GEF is' cut by the 
three parallels, be given ; the third parallel HK is given in 
position. 

In AB take a given point L, and draw LM perpendicular 
to CD, meeting HK in N ; because LM is drawn from the 
given point L to CD which is given in position j and makes 
a given angle LMD ; LM is given in position' j and CD is * i3. dit. 
given in position, wherefore the point Al is given" ; and the " 23.d2t.' 
point L is given, LM is therefore given in magnitude'^ ; and ecg.dat. 
because the ratio of GE to GF is given, and as GE to GP , so 

H G N K A E L B 

/" I 

L B H IG "Si K 




( Cor. 

is NL to NM ; the ratio of NL to NM is given; and therefore* ) ^' "' 
the ratio of ML to LN is given ; but LM is given in magni- 
tude'', v/herefore« LM is given in magnitude : And it is also " 2 dat. 
given in position, and the point L is given, wherefore^ the f^o. dat. 
point N is given, and because the straight line HK is drawn 
through the given point N parallel to CD, which is given in 
position, therefore HK is given in positions. 1 31 da:. 



PROP. XLL F. 

J F a straight line meets three parallel straight lines s^^^"- 
which are givea in position, the segments into 
which they cut it have a given ratio. 

Let the parallel straight lines AB, CD, EF given in posi- 
tion, be cut by the straight lineGHK j the ratio of GH tg HK 
is given. 

In A B take a given point L, and J^ G ] 15 
draw LM perpendicular to CD, meet- T' "T 

ing EF in N j therefore^ LM is given Q TT/ ^Fi T) 

in position J and CD, EF are given "" "" ^ » o. !: 

in position, wherefore the points M, 

N are given: And the point Lis given; ^^-^ '^ 

therefore*' the straight lines LMjMN' -t- K. . N F ,. ^^^. 
are given in magnitude ; and the ratio 

C c 2 



388 EUCLID'S 

M.dat. of LMtoMN is therefore given^: But as LM to MN, s# 
is GH to HK J wherefore the ratio of GH to HK is given. 



39. 



PROP. XLII. 



See Jf . 



IF each of the sides of a triangle be given in mag- 
nitude, the triangle is given in species. 

Let each of the sides of the triangle ABC be given in mag- 
nitude, the triangle ABC is given in species. 
• 52. 1. Make a triangle^ DEF, the sides of which are equal, each 

to each, to the given straight lines' AB, BC, CA, which can 
be done ; because any two of them must be greater than the 
third ; and let DE be 

equal to AB, EF, to BC, \ jy 

and FD to CA ; and be- 
cause the two sides ED, 
DF are equal to the two 
BA, AC, each to each, 
and the base EF equal to 
the base BC ; the angle 

"8.1. EDF is equally to the angle BAG; therefore, because the 
angle EDF, which is equal to the angle BAC, has been found, 

c \. def. ^^^ angle BAC is given'', in like manner the angles at B, C, 
are given. And because the sides AB, BC, CA are |iven^ 

* 1. dat. their ratios to one another are given'', therefore the triangle 

"5. dcf. ABC is given^ in species. » 





-iO, 



PROP. XLIIL 

J.F each of the angles of a triangle be given in 
magnitude, the triangle is given in species. 

Let each of the aiigles of the triangle ABC be given in 
magnitude, the triangle ABC is given in species. 

Takeastraightline DEgivenin 
position and magnitude, and at the 
points D, E make^* the angle EDF 
equal to the angle BAC, and the 
angle DEF equal to ABC; there- , 

fore the other angles EFD, BCA ■" ^' I:^ 

are equal, and each of the angles at the points A, B, C, is 

I given 




data: 389^ 

given, wherefore each of those at the points D, E, F is given : 
And because the straight line FD is drawn to the given point 
D in DE which is given in position, making the given angle 
EDF ; therefore DF is given in position*'. In like manner »32.dai. 
EF also is given in position ; wherefore the point F is given : 
And the points D, E are given ; therefore each of the straight 
lines DE, EF, JtD is given"^ in magnitude; wherefore the ^ 29. dar. 
triangle DEF is given in species'*: and it is similar* to the <• 42.dat. 
triangle ABC; which therefore is given in species. «/t.'dk. 

( «• 

PROP. XLIV. 41. 

J[F one of the angles of a triangle be given, and if 
the sides about it have a given ratio to one another; 
the triangle is given in species. 

Let the triangle ABC have one of its angles BAG given, 
and let the sides BA, AC about it have a given ratio to one 
another ; the triangle ABC is given in species. 

Take a straijjht line DE given in position and magnitude, 
and at the point D, in the given straight line DE, make the 
angle EDF equal to the given angle BAG ; wherefore the 
angle EDF is given ; and because the straight line FD is drawn 
to the given point D in ED which is given in position, making 
the given angle EDF ; therefore * 

FD is given in position\ And . ^ Tv,32.d.t. 

because the ratio or BA to AV^ is y^ ^ 

given, make the ratio of ED to y^ y^ 

DF the same with it, and join EF; /^ 1^ y^ - 1 

and because the ratio of ED to DF R C K F 

is given, and ED is given, therefore^ DF is given in magni- "2. da?. 
tude : and it is given also in position, and the point D is given, 
wherefore the point F is given'': and the points D, E are^^***** 
given, wherefore DE, EF, FD are given*^ in magnitude : * ^' t^af. 
and the triangle DEF is therefore given* in species ; and be- «42.d»«. 
cause the triangles ABC, DEF have one angle BAC equal to 
oneangle EDF, and the sides about these angles proportionals ; 
the triangles are*" similar ; but the triangle DEF is given in *^6'*' 
species, and therefore also the triangle ABC. 

Cc3 



3^0 



EUCLID'S 



See N. 



» 2. dat. 



«> 22. 5. 



<20. 1. 



OA. 5. 



f 22. 1. 



f 42. dat. 



6 5.6. 



42. PROP. XLV. 

IF tlie sides of a triangle have to one another 
given ratios; the triangle is given in species. 

Let the sides of the triangle ABC have given ratios to one 
another, the triangle ABC is given in species. 

Take a straight line D given in magnitude ; and because 
the ratio of AB to BC is given, make the ratio of D to E the 
.same with it; and D is given, therefore^ E is given. And 
because the ratio of BC to CA is given, to this make the ratio 
of E to F the same ; andE is given, and therefore^ F. And 
because as AB to BC, so is D to E i by composition AB and 
BC^together are to BC, as D 
anii E to E ; but as BC to CA, 
so is E to F ; therefore, ex 
asquali^, as AB and BC are to 
CA, so are D and E to F, and 
AB and BC are greater*^ than 
CA ; therefore D and E are 
greater"* than F. In the same 
manner any two of the three D 
E, F are greater than thethird. 
Make ^ the triangb GHK 

whose sides are equal to D, E, F, so that GH be equal to D, 
HK to E, and KG to F ; and because D, E, F, are, each of 
them, given, therefore GH, HK, KG are each of them given 
-in magnitude ; therefore the triangle GHK is given*' in spe- 
cies : But as AB to BC, so is (D to E, that is) GH to HK ; 
and as BC to CA, so is (E to F, that is) HK to KG j there- 
fore, ex aequali, as AB to AC, so is GH to GK. Where- 
fores the triangle ABC is equiangular and similar to the tri- 
angle GHK ; and the triangle GHK is given in species ; 
therefore also the triangle ABC is given in species. 

CoR. If a triangle is required to be made, the sides of 
which shall have the same rati<bs which three given straight 
lines D, E, F have to one another ; it is necessary that every 
two of them be greater than the third. 




D E V 



DATA. 391 

PROP. XLVI. 43. 

i-F the sides of a right angled triangle about one 
of the acute angles have a given ratio to one ano- 
ther ; the triangle is given in species. 

Let the sides AB, BC about the acute angle ABC of the 
triangle ABC, which has a right angle at A, have a given ra- 
tio to one another j the triangle ABC is given in species. 

Take a straight line DE given in position and magnitude ; 
and because the ratio of AB to BC is given, make as AB to 
BC, so DE to EF J and because DE has a given ratio to EF, 
and DE is given, therefore* EF is given ; and because as AB '2. dat. 
to BC, so is DE to EF -, and AB ts less^ than BC, therefore b 19. i. 
DE is less' than EF. From the point D draw DG at right an- «^ a-5. ^ 
gles to DE, and from the cen- 
tre E, af the distance EF, \ 
describe a circle which shall ii>- JlX^"" 
meet DG in two points j let -n ^^ \ ^ 
G be either of them, and ^-* ^ 
join EG ; therefore the cir- 
cumference of the circle is 
given<^ in position; and the straight line DG is given ^ in^^'^f"- 
position, because it is drawn to the given point D in DE given "^ ''-•*'*■• 
in position, in a given angle ; therefore*" the point G is given, f28.dat. 
and the points D, E are given, wherefore DE, EG, GD areeap. dat. 
givens in magnitude, and the triangle DEG in species^ >» 42. d.;. 
And because Che triangles ABC, DEG have the angle BAG 
equal to the angle EDG, and the sides about the angles ABC, 
DEG proportionals, and each of the other angles BCA, EGD 
less than a right angle ; the triangle ABC is equiangular' and i -. -. 
similar to the triangle DEG; but DEG is given in species; 
therefore the triangle ABC is given in species: And in the 
same manner, the triangle made by drawing a straight line 
irom E to the other point in which the circle meets DG is 
given in species. 

C c 4 




39^ 



E U C L 1 D'S 



W- 



PROP. XLVir. 



Pee N. 



»32. 1. 
b 43. «iat. 



^ 32. dat. 



1 2. dat. 

*A. 5: 



fe. dcf. 


e28. 


dat. 


h 29^ 


dat, 


i42. 


dat. 


k]8, 


. 1. 


M7. 


1. 




If a triangle has one of its angles which is not a- 
right angle given, and if the sides about another 
angle have a given ratio to one another; the trian- 
gle is given in species. 

Let the triangle ABC have one of its angles ABC a given, 
but not a right angle, and let the sides B A, AC about another 
angle BAG have a given ratio to one another j the triangle 
ABC is given in species. 

First, Let the given ratio be the ratio of 
equality, that is, let the sides BA, AC, and 
consequently the angles ABC, ACB, be 
equal j and beeau.se the angle ABC is given, 
the angle ACB, and also the remaining* angle 
BAC is given ; therefore the triangle ABC is 
given^ in species ; and it is evident that in this 
case the given angle ABC must be acute. 

Next, Let the given ratio be the ratio of a less to a greater, 
that is, let the side AB adjacent to the given angle be less 
than the side AC : Take a straight line DE given in position 
and magnitude, and make the angle DEF equal to the given 
angle ABC; therefore EF is given'^ in position i and because 
the ratio of BA to AC is given, 
as BA to AC, so make ED to 
DG ; and because the ratio of 
ED to DG is given, and ED is 
given, the straight line DG is 
given"*, and BA is less than 
AC, therefore ED is less ^ than 
DG. From the centre D, at 
the distance DG, describe the 
circle GE, meeting EF in F, 
and join DFj and because the 
circle is given ^ in position, 
as also the straight line EF, the 
point F is given e ; and the 
points D, E are given ; where- 
fore the straight Imes, DE, EF, FD are given •* in mag- 
nitude, and the triangle DEF in species'. And because BA 
is less than AC, the angle ACB is less'' than the angle ABC, 
and therefore ACB is less' than a- right angle. In the same 

manner, 





DATA. 

manner, because-ED is less than DG or DF, the angle DFE 
is less than a right angle: And because the triangles ABC, 
DEF have the angle ABC equal to the angle DEF, and the 
sides about the angles BAG, EDF proportionals, and each 
of the other angles ACB, DFE less than a right angle ; the 
triangles ABC, DEF are == similar, and J3EF is given in spe- 
cies, wherefore the triangle ABC is also given in species. 

Thirdly, Let the given ratio be the ratio of a greater to a 
less, that is, let the said AB adjacent to the given angle be 
greater ihan AC ; and as in the last 
case, take a straight line DE given in 
position and magnitude, and make the 
angle DEF equal to the given angle 
ABC ; therefore EF is given^ in posi- 
tion : Also draw DG perpendicular to 
EF ; therefore if the ratio ofBA to 
AC be the same with the ratio of ED 
to the perpendicular DG, the triangles 
ABC, DEG are similar^, because the 
angles ABC, DEG are equal, and 
DGE is a right angle : Therefore the 
angle ACB is a right angle, and the triangle ABC is given 
in° species. 

But if, in this last case, the given ratio of BA to AC be not 
the same with the ratio of ED to DGy that is, with the 
ratio of BA to the perpendicular AM drawn from A to BC ; 
the ratio of BA to AC must be less than " the ratio of BA 
to AM, because AC is greater than AM. Make as B A to AC, 
so ED to DH ; therefore the ratio of 
ED to DH is less than the ratio of ( B A 
to AM, that is, than the ratio of) ED 
to DG ; and consequently DH is greater? 
than DG ; and because BA is greater 
than AC, ED is greater' than DH. 
From the centre D, at the distance DH, 
describe the circle KHF which necessa- 
rily meets the straight line EF in two 
points, because DH is greater than DG, 
and less than DE. Let the circle meet 
EF in the points F, K which are given, 
as was shewn in the preceding case ; and DF ; DK. being joined, 
the triangles DEF, DEK are given in species, as was there 

shewn, 



393 



°7,6. 



« 32. (Ut. 



* 43.dat. 




394 EUCLID'S 

shewn; From the centre A, at the distance AC, describe a 
circle meeting BC again in L : And if the angle ACB be less 
than a right angle, ALB must be greater than a right angle ; 
and on the contrary. In the same manner, if the angle DFE 
be less than a right angle, DKE must be greater than one j 
and on the contrar/U Let each of the 
angles ACB, DFL be either less or 
greater than a right angle ; and because 
in the triangles ABC, DtF the angles 
ABC, DEF are equal, and the sides 
BA, AC, and ED, DF, about two of 
the other angles proportionals, the tri- 
""T, 6. angle ABC is similar "" to the trianglp 
«DEF. Jn the same manner, the tri- 
angle ABL is similar to DEK, And 
the triangles DEF, DEK are given 
in species ; therefore also the triangles 
ABC, ABL are given in species. And from this it is evi- 
dent, that, in this third case, there are always two triangles 
of a different species, to which the things mentioned as given 
in the proposition* can agree. 




* 9. i. 
«> 3. 6. 



« 12. 5. 



" 47. daf. 
« 43, daf. 



-,. PROP. XLVIIL 

If a triangle has one angle given, and irboth the 
sides tooether about that an^-le liave a e'iven ratio to 
the remaining side ; the triangle is given in species. 

Let the triangle ABC have the angle BAC given, and let 
the sides BA, AC together about that angle have a given,ratio 
to BC ; the triangle ABC is givdn in species. 

Bisedl^ the angle BAC by the straight line AD j therefore 
the angle BAD is given. -And because as B A to AC, so is^ 
BD to DC, by permutation, as AB to BD, 
8o is AC to CD ; and as BA and AC to- A 

gether to BC, so is'^ AB to B D. But the 
ratio of BA and AC together to BC is 
given, wherefore the ratio of AB to BD ^ 
is given, and the angle BAD is given ; J^ 
therefore ^ the triangle ABD is given in 
species, and the angle ABD is therefore given ; the angle BAC 
is also given, wherefore the triangle ABC is given in«pecies . 

A triangle which shall have the things that are mentioned 
in the proposition to be given, can be found in the following 

manner. 




D A T- ^A*^ i. 395 

manner. Let EFG be the given angle, and let the ratio of H 
to K be the given ratio which the two sides about the angle 
EFG must have to the third side of the triangle ; therefore be- 
cause two sides of a triangle are greater than the third side, 
the ratio of H to K must be the ratio of a greater to a less. 
BisecT the angle EFG by the straight line FL, and bv the '•- ^' 
47th proposition find a triangle of which EPL is one of the 
angles, and in which the ratio of the sides about the angle op- 
posite to FL is the same with the ratio of H toK: Todo which, 
take FE given in position and magnitude, and draw EL per- 
pendicular to FL : Then if the ratio of H to K be the same 
with the ratio of FE to EL, produce EL, and let it meet FG 
in P ; the triangle FEP is that which was to be found : For it 
has the given angle EFG ; and 
because this angle is bisefted by 
FL, the sides EF, FP together 
are to EP, as'' FE to EL, that 
i , asFItoK. 

But if the ratio of H to K 
be not the same with the ratio 
of FE to EL, it must be less 

than it, a^ was shewn in Prop. 47, and in this case there are two 
triangles, each of which has the given an^e EFL, and the 
ratio of the sides about the angle opposite to FL the same with 
the ratio of H to K. By Prop. 47, find these triangles EFM, 
' EFN, each of which has the angle EFL for one of its angles, 
and the ratio of the side FE to EM or EN the same with the 
ratio of H to K ; and let the angle EMF be greater, and ENF 
less than a right angle. And because H is greater than K, EF 
is greater than EN, and therefore the angle EFN, that is, the 
angle NFG, is less^ than the angle ENF. To each of these f 13. 1. 
add the angles NEF, EFN ; therefore the angles NEF, EFG 
are less than the angles NEF, .EFN, FNE, that is, than two 
right ajigles j theretore the straight lines EN, FG must meec 
together when produced ; let them meet in O, and produce: 
Ei\I to G. Each of the triangles EFG, EFG has tke things 
mentioned to be given in the proposition : For each of them 
has the given angle EFG ; and because this angle is bisected 
by the straight line FAIN, the sides EF,FG together have to 
EG the third side the ratio of FE to EM, that is, of H to K. 
In like manner, the sides EF, FO together have to EG the 
ratio which H has to K. 




396 E U C L I D'S 

w. • PROP. XLIX. 

If a triangle has one angle given, and if the sides 
about another angle, both together have a given 
ratio to the third side ; the triangle is given in species. 

Let the triansfle ABC have one angle ABC given, and let 
the two sides B A, AC about another angle B AC have a given 
ratio to BC ; the triangle ABC is giv^en in species. 

Suppose the angle BAC to be bisedlcd by the straight line 
AD ; BA and AC together are to BC, as AB to BD, as was 
shewn in the preceding proposition, ^ut the ratio of BA and 
AC together to BC is given J therefore also the ratio of A B to 
"44. dat. BD is given. And the angle ABD is given, wherefore* the 
triangle ABD is given in species ; and consequently the angle 
BAD; and its double the angle BAC 
are given : And the angle ABC is 
given. Therefore the triangle ABC 
* 43. dat. is given in species^. 

A triangle which shall have the 
things mentioned in the proposition 
to be given, may be thus found. Let 

EFG be the given angle, and the ratio -rr E 

ofH to K the given ratio; and by -tr ^-'^'/XllT 

. Prop. 44. find the triangle EFL, ^ ^^ I v'" 

which has the angle EFG for one of ^ / \ 

its angles, and the ratio of the sides p Tj O 

EF, FL about this angle the same 

with the ratio of H to K ; and make the angle LEM equal to 
the angle FEL. And because the ratio of H to K is the ratio 
which two sides of a triangle have to the third, H must be 
greater than K : and because EF is to FL, as H to K ; there- 
fore EF is greater than FL, and the angle FEL, that is, LEM, 
is therefore less than the angle EIE. Wherefore the angles 
LFE, P'EM are less than two right angles, as was shewn in 
the foregoing proposition, and the straight lines FL, EM muat^ 
meet if produced ; let them meet in G, EFG is the triangle 
which was to be found ; for EFG is one of its angles, and be- 
cause the angle FEG is bjsedled. by EL, the two sides FE, 
FXj together have to the third side FG the ratio of EF to FL, 
that is, the given ratio of H to K. 





D ATA. 397 

PROP. L, T(i. 

IF from the vertex of a triangle given in species, a 
straight line be drawn to the base in a given angle ; 
it shall have a given ratio to the base. 

From the vertex A of the triangle ABC which is given ia 
species, let AD be drawn to the base BC in a given angle 
ADB j the ratio of AD to BC is given. 

Because the triangle ABC is given in 
species, the angle ABD is given, and the 
angle ADB is given, therefore the tri- 
angle ABD is given ' hi species ; where- / V ^V^^^ **3. dn. 
fore the ratio of AD to AB is given. 
And the ratio of AB to BC is given ; and 
therefore" the ratio of AD to BC is given. » 9. da;. 

PROP. LI. M. 

AXecti LINEAL figurcs given in species, are dir 
vided into triangles which are given in species. 

Let the re£lilineal figure ABCDE be given in species : 
ABCDE may be divided into triangles given in species. 

Join BE, BD ; and because ABCDE is given in species, 
the angle B AE is given% and the ratio jV » 3. d?f. 

ef BA to AE is given*; wherefore 
the triangle BAE is given in species", 
and the angle AEB is therefore given'. 
( But the whole angle AED is given, 
and therefore the remaining angle BED 
is given, and the ratio of AE to EB 
is given, as also the ratio AE to ED ; therefore the ratio of 
BE to ED is given^. And the angle BED is given, where- ' 
fore the triangle BED is given** in species. In the same man- 
ner, the triangle BDC is given in species : Therefore rc^l- 
iineal figures which are given in species are divided into, trio" 
angles given in species. 




398 EUCLID'S 

43. PROP. Lll. 

If two triangles given in species be described upon 
the same straight hne ; they shall have a given ra- 
tio to one another. 

Let the triangles ABC, ABD given In species be described 
upon the same straight line AB ; the ratio of the triangle ABC 
to the triangle ABD is given. 

Through the point C, draw GE parallel to AB, and let it 
meet DA produced in E, and join BE. Because the triangle 
ABC IS given in species, the angle BAG, that is, the an*;le 
ACE, is given; and because the triangk ABD is given in 
species, the angle 



?. 1. 



DA B, that is, the El__ ■. C 




angle AEC, is given. 

XherefoFe the trian- 
gle ACE is given in. 

species ; wherefore the. 

ratio of EA to AC is 
» 3. clef. given% and the ratio 

of CA to AB is given, 
k 9. riat. as also the ratio of BA to AD ; therefore the ratio of b E A to 
■^21. 1. AD is given, and the triangle ACB is equa^ to the triangle 

AEB, and as the triangle AEB, or ACB, is to the triangle 
•1 !. P. ADB, so is ^ the straight line EA to , AD : But the ratio oi'" 

EA to AD is given ; therefore the ratio of the triangle ACB 

to the triangle .ApB is given. 

■^-.;. .'m1^-'' PROfiLEM; , ,,. 

To firii^ (he ratio of two triangles ASCj ABI) given in spe- 
cies, and which-are described upon the same straight line Al>. 

Take a straight line FG given in position and magnitude, 
and because the angles of the triangles ABC, ABD are given, 
at the points F, G of the straight line FG, make the angles 
GFH, GFK"= equal to the angles BAG, BAD ; and the angles 
FGH, FGK equal to the angles ABC, ABD, each to each. 
Therefore the triangles ABC, ABD are equiangular to the tri- 
angles FGII, FGK, each, to each. Through the point H draw 
HL parallel to FG, meeting KF produced in L. And because 
theungles BAG, BAD are equal to the angles GFH,GFK, each 
to each ; therefore the angles v\GE, AEC arc equal to FHL, 
FLH, each to each, and the triangle AEC equiangular to the 
triangle FLH. Therefore as EA to AC, so is LF to FH, and 

as 



DATA. 399 

as CA to AB, so HF to FG j and as BA to AD, so is GF 
toFK ; wherefore, ex Equali, as EA to AD, so is LF to FK. 
But as was shewn, the triangle ABC is to the triangle ABD, 
as the straight line EA to AD, that is, as LF to FK. The 
ratio therefore of LF to FK has been fpaud, which is the same 
with the ratio of the triangle ABC to thCjCri^Dgle ABD. 

PROP. LIII. ? 

If two rectilineal figures given in species. be de-sesx. 
scribed upon the same strais:bt line ; they shall 
have a given ratio to one another. 

Let any two rectilineal figures ABCD.E,ABFG, which are 
given in species, be described upon the .same st^ight iijie AB i 
the ratio of them to one another is given. ^ ... $ / . . 

Join AC, AD, AF ; each of the triangles AED^ ADC, 
ACB, AGF, ABF is gi ven^ in species. And b^cau£« the t^- * Ji. dat. 
angles ADF, ADC given in spe- :^ > 

cies dro described upon the same D 

straight line AD, the ratio of EAD F^-^-""/^ \^ 
to DAC is given'' j and, by com- 't"^^ / S^^ " *-• °^^* 

position, the ratio of EACD to \ X ^^^-^''^ j 

DAC is given':. And the ratio -Aj^:;;;____/t» ' "• ''''^• 
DAC to CAB is given"*, because 7^ \ 

they are described upon the same Q.L ^~"~~''~~Ay 

straight line AC ; therefore the . 

ratio of EACD to ACB is given"^; | £ K L jfN q 

and, bv composition, the ratio of 

ABCDE to ABC is given. In the same manner>ther4tioor 

ABFG to ABF is given. But the ratio of the triangle ABC 

to the triangle ABF is given ; wherefore^ because the ratio of 

ABCDE to ABC is given, as also the ratio of ABC to ABF, 

and the ratio of ABF to ABFG ; the ratio of the redilioeai 

ABCDE to the rectilineal ABFG is given*^. 

PROBLEx\L 

To find the ratio of two rectilineal figures given in species, 
and described upon the same straight line. 

Let ABCDE, ABFG be two rectilineal figures given in 
species, and described upon the same straight line AB, and 
join AC, AD, AF, Take a straight line HK given in posicioti 
and magnitude, and by the 52d dat. find the ratio of the tri- 
angle ADE to the triangle ADC, andmalft the ratio of HK 



400 £ U C L I D'S 

to KL the same with it. Find also the ratio of the triangle 
ACD to the triangle ACB. And make the ratio of KL to 
LM the same. Also, find the ratio of the triangle ABC to 
the triangle ABF, and make the ratio of LM to MN the sartie* 
And, lastly, find the ratio of the triangle AFB to the triangle. 
AFG, and make the ratio of MN 
to NO the same. Then the ratio 
of ABCDE to ABFG is the same 
with the ratio of HM to MO. 

Because the triangle £AD is to 
the triangle DAC, as the straight 
line HK to KL ; and as the triangle 
DAC to CAB, so is the straight 
line KL to LM ; therefore by using ^ _ nir-Kr 

composition as often as the number Jj ^ A- X^ ^^^ Q 
of triangles requires, the redtilineal 

ABCDE is to the triangle ABC, as the straight line HM to 
ML, In like- manner because the triangle GAF is to FAB, 
as ON to NM, by composition, the re6iiiineal ABFG is to 
the triangle ABF as MO to NM, and by inversion, as ABF 
to ABFG, so is NM to MO. And the' triangle ABC is to 
ABF, as LM to MN. Wherefore, because as ABCDE to 
ABC, so is HM to ML ; and as ABC to ABF, so is LM to 
MN ; and as ABF to ABFG, so is MN to MO ; ex iequali, 
as the redilincal ABCDE to ABFG, so is the straight HM 
to MO. 




.=-0. 



PROP. LIV. 



XF two straight lines have a given patio to one ano* 

ther ; the similar rectilineal figures described upon 
them similarly, shall have a giv- en ratio to one another. 

Let the straight lines AB, CD, have a given ratio to one an- 
other, and let the similar and similarly placed reftilineal figures 
E, F be described upon them j the ratio of E to F is given. 

ToAB, CD, let G be a third 
proportional ; therefore as AB to 
CD, so is CD to G. And the ratio 
of AB to CD is given; wherefore 
the ratio of CD to G is given ; and 
consequently the ratio of AB to G 

.^ (lat. is also given*. But as A B to G, so 

so.cT'"^* '^ ^^^ figure E to the figure^ F. Therefore the ratio of E to 
F is given. 




DATA. 

PROBLEM. 



401 



20.6. 



To find the ratio of two similar rectilineal figures E, F, simi- 
larly described upon straight lines AB,CD which have a given 
ratio to one another ; Let G be a third proportional to AB,CD. 

Take a straight line H given in magnitude ; and because 
the ratio of AB to CD is given, make the ratio of H to K 
the same with it ; and because H is given, K is given. As 
H is to K, so make K to L ; then the ratio of E to F is the 
same with the ratio of H to L ; for AB is to CD, as H to K, 
wherefore CD is to G, as K to L ; and, ex aequali, as AB to 
G, so is H to L : But the figure E is to'' the figure F, as AB *> -^^<^ 
to G, that is, as H to L. 

PROP. LV. 

If two straight lines have a given ratio to one ano- 
ther ; the rectilineal figures given in species, described 
upon tliem, shall have to one another a given ratio. 

Let AB, CD be two straight lines which have a given ratio 
to one another ; the rectilineal figures E, F given in species 
and described upon them, have a given ratio to one another. 

L^pon the straight line AB, describe the figure AG similar 
and similarly placed to the figur* F ; and because F is given in 
species, AG is also given in spe- 
Therefore, since the fi- 



H- 



B 

Gr 

Kr 



C D 



gures E, AG which are given A 

in species, are described upon 

the same straight line AB, the 

ratio of E to AG is given% 

and because the ratio of AB to 

CD is given, and upon them are described the similar and 

similarly placed rediiineal figures AG, F, the ratio of AG to b54.;dat 



L 



a 33. dat. 



is given i therefore 



F is given'' j and the ratio of AG to E 
the ratio of E to F i* given'. 

PROBLEM. 

To find the ratio of two redilineal fitrures E, F griven in 
species ^nd described upon the straight Imes AB, CD which 
have a given ratio to one another. 

Take a straight line H given in magnitude; and because 
the recftilineal figures E, AG given in species, are described up- 
on the same straight line AB, find their ratio by the 53d dat. 
and make the ratio of H to K the same, K is therefore given. 
And because the similar rectilineal figures AG, F are described 

D d upon 



« 9. dat. 



401 EUCLID'S 

upon the straight lines AB, CD, which have a given ratio, 
find their ratio by the 54th dat. and make the ratio of K to L 
the same ; The figure E has to F the same ratio which H has 
to L : For, by the eonsti uftion, as E is to AG, so is H to K ; 
and as AG to F, so is K to L ; Therefore, ex sequali, as E 
to F i so is H to L. 



b2. 



1 



PROP. LVI. 



>F a rectilineal figure given in specves be described 
upan a straight line given in ,niagnitude; the fi- 
gure is given in magnitude. 

Let the re£tilineal figure ABODE given in species, be de- 
scribed upon the straight line AB given in magnitude j the 
figure ABCDE is given in magnitude. 

Upon AB let the square AF be described ; therefore AF is 

given in species and magnitude, and because the rectilineal 

figures ABCDE, AF given in species 

are described upon the same straight 

line AB, the ratio of ABCDE to AF is 

*53. dat. given* : But the square AF is given in 

2. dat, maj;;nitude, therefore^ also the figure 

ABCDE is given in magnitude. 

PROB. 

Tofindthemagnitudeofaredlilineal 
figure given in species described upon a 
straight line given in magnitude. 

Take the straight line GH equal to 
the given straight line AB, and by the 
53ddat. find the ratio which the square 
AF upon AB has to the figure AtsCDE ; and make the ratio 
of GH to HK the same i and upon GH describe tfiesquare GL, 
and complete the parallelogram LHKM ; thp fipiurc ABCDE 
is equal to LHKM j because AF is to ABCDE, as the straight 
line GH to HK,that is, as the figure GL to HM ; and AF is 
equal to GL > therefore ABCDE is equal to HM^ 

PROP. LVIL 

IF two rectilineal figures arc given in species, and 
if a side of one of them has a given ratio to a side 
of the other ; the ratios of the remaining sides to the 
remaining sides shall be given. 

Let 




«:]*. 5 



DATA. 



403 



Let AC, DF be two re<£lilineal figures given In species, and 
let the ratio of the side AB to the side DE be given, the ratios 
of the remaining sides to the remaining sides are also given. 

Because the ratio of AB to DE is given, as also= the ratios »3. deO 
of AB to BC, and of DE to EF, the ratio of BC to EF is 
given^ In the same manner the * lO- da»- 

ratios of the other sides to the 
other sides are given. 

The ratio which BC has to 
EF may be found thus : Take a 
straight line G given in magni- 
tude, and because the ratio of BC 
to BA is given, make the ratio 
of G to H the same ; and because 
the ratio of AB to DE is given, 
make the ratio of H to K the 

same ; and make the ratio of K to L the same with the given 
ratio of DE to EF. Since therefore as BC to BA, so is G to 
H ; and as BA to DE, so is H to K ; and as DE to EF so is^ 
K to L J ex iequali, BC is to EF, as G to L j therefore the 
ratio of G to L has been found, which is the same with the 
ratio of BC to EF. 




PROP. LVIII. 



c. 



2 Cor. 
20.6, 



J F two similar rectilineal figures have a given ratio ^*^* 
lO one another, tlieir homologous sides have also a 

triven ratio to one another. 

o 

Let the two similar r&Shlineal figures A, B have a given 
ratio to one another, their homologous sides have also a given 
ratio. 

Let the side CD be homologous to EF, and to CD, EF let 
the straight line G be a third proportional. As therefore* CD » 
to G, so is the figure A to B ; and 
the ratio of A to B is given, there- 
fore the ratio of CD to G is given j 
and CD, EF, Gare proportionals ; 
wherefore'' the ratio of CD to EF 
is given. 

The ratio of CD to EF may be 
found thus : Take a straight line 

H given in magnitude ; aiid because the ratio of the figure A 
to B is given, make the ratio of H to K the same with it : 
And, as the 13th dat. direds to be done, find a mean propor- 

D d 2 tional 




1^ * 13. iiu 



H 



L IC 



404 EUCLID'S 

tional L between H and K j the ratio of CD to EF is the same 
with that of H to L. Let G be a third proportional to CD, 
EF i therefore as CD to G, so is (A to B, and so is) H to K ; 
and as CD to EF, so is H to L, as is shewn in the 13th dat. 

54. PROP. LIX. 

Se«N. If two rectilineal figures given in species have a 
given ratio to one another, their sides shall likewise 
have given ratios to one another. 

Let the two reftilineal figures A, B, given in species, have 
a given ratio to one another, their sides shall also have given 
ratios to one another. 

If the figure A be similar to B, their homologous sides shall 
have a given ratio to one another, by the preceding proposi- 
tion ; and because the figures are ;^iven in species, the sides of 
• 3. dif. c^^h of them have given ratios* to one another ; therefore each 
" 9, dati side of two of them has*" to each side of the other a given ratio. 
But if the figure A be not similar to B, let CD, EF be 
any two of their sides; and upon EF conceive the figure EG 
to be described similar and 
similarly placed to the figure 
A, so that CD, EF be hu- 



mologous sides j therefore C -T) £ N. B / •*' 
EG is given in species ; and — 

the figure B is given in spe- W "^ 



K 



« 53. dat. cjgs . vvherefore = the ratio 

of B to EG is given j and the -ji^ 
ratio of A to B is given, -y^ 
therefore ^ the ratio of the L 
figure A to EG is given ; and A is similar to EG ; therefore 

"58. dat. ''the ratio of the side CD to EF is given ; and consequently'' 
the ratios of the remaining sides to the remaining sides are 
given. 

The ratio of CD to EF may be found thus: Take a straight 
line H given in magnitude, and because the ratio of the figure 
A to B is given, make the ratio of H to K the same with it. 
And by the 53d dat. find the ratio of the figure B to EG, and 
make the ratio of K to L the same : Between H and L find a 
mean proportional M, the ratio of CD to EF is the same with 
the ratio of H to Mj because the figure A is to B as H to K j 
aind as B to EG, so is K to L j ex %quali, as A to EG, so is 

H tp 



DATA. 

H to L : And t?ne figures A, EG, are similar, and M is a mean 
proportional between H and L j therefore, as was shewn in 
the preceding proposition, CD is to EF as H to M. 



405 



PROP. LX. 



55, 




K 




If a rectilijieal figure be given in species and mag^ 
nitude, the sides of it shall be given in magnitude. 

Let the redilineal figure A be given in species and magni- 
tude, its sides are given in magnitude. 

Take a straight line BC given in position and magnitude, 
and upon BC describe* the figure D similar, and similar!)' * ^8. 6. 
placed, to the figure A, 
and let EF be the side 
of the figure A homo- 
logous to BC the side 
of D; therefore the fi- 
gure D is given in spe- 
cies. And because upon 
the given straight line 
BC the figure D given 
in species is described, D 
is given'' in magnitude, and the figure A is given in magni- ^ tg j^( 
tude, therefore the ratio of A to D is given : And the figure 
A is similar to D ; therefore the ratio of the side EF to the 
homologous side BC is given'^ ; and BC is given, wherefore'* 1 2%au* 
EF is given : And the ratio of EF to EG is given', therefore • 3. def. 
EG is given. And, in the same manner, each of the other 
sides of the figure A can be shewn t* be given. 

PROBLEM. 

To describe a reftilineal figure A similar to a given figure D, 
and equal to another given figure H. It is Prop. 25, B. 6.Elem. 

Because each of the figures D, H is given, their ratio is gi- 
ven, which may be found by making^ upon the given straight fcor. 45. 1. 
line BC the parallelogram BK equal to D, and upon its side 
CK making^ the parallelogram KL equvil to H in the angle 
KCL equal to the angle MBC ; therefore the ratio of D to H, 
that is, of BK to KL, is the same with the ratio of BC to CL : 
And because the figures D, A are similar, and that the ratio of 
D to A, or H, is the same with the ratio of BC to CL; by 
the 58th dat. the ratio of the homologous sides BC, EF is the 
same with the ratio of BC to the mean proportional between 
BC and CL. Find EF the mean proportional j then EF is the 

D d 3 side 



4o6 EUCLID'S 

side of the figure to be described, homologous to BC the side 

of D, and the figure itself can be described by the i8th Prop. 

B. 6, which, by the construction, is similar to D ; and because 

6 2. cor. D is to A, as 8 BC to CL, that is, as the figure BK to KL ; and 

^ 20. 6. jhat D is equal to BK, therefore A'' is equal to KL, that is, to H, 

57. PROP. LXI. 

siceK. JLF a parallelogram given in magnitude has one of 
its sides and one of its angles given in magnitude, 
the other side also is given. 

Let the parallelogram ABDC given in magnitude, have the 
side AB and the angle BAC given in magnitude, the other 
side AC is given. 

Take a straight line EF given in position and magnitude j 
and because the parallelogram AD 
is given in magnitude, a redtilincal 

»i.def. figure equal to it can be found*. 
And a parallelogram equal to this 

»^<3r 45. 1 figure can be applied^ to the given 
straight line EF in an angle equal 
to the given angle BAC. Let this 
be the parallelogram EFHG, hav- 
ing the angle FEG equal to the 

angle BAC. And because the pa- Q J{_ 

rallel' g ams AD, EH are equal, and 

have tne angles at A and E equal ; the sides about them ar<e 

c 14 g. reciprocally proportional'^ ; therefore as AB to EF, so is EG 
to AC : And AB, EF, EG are given, therefore also AC is 

d j2. 6. given''. Whence the way of finding AC is manifest. 



A 


r. 


/ 


/ 


C 


.]) 


; / 



H. PROP. Lxn. 

»e« N. If a parallelogram has a given angle, the rectangle 
contained by the sides about that angle has a given 
ratio to the parallelogram. 

Let the parallelogram ABCD have the 
given angle ABC, the re«^ngle AB, 'BC 
has a given ratio to the parallelogram AC. 
From the point A draw AE perpendi- 
cular to BC; because the angle ABC is 
given, as also the angle AEB, the triangle 
••43.dat. ABE is given* in species ; therefore the 
ft^tio of BA to AE is given. But as BA to 
AE, so is*" the reAangle AB, BC, to the 
rectangle AE, BC, therefore the ratio of 



I 




DATA. 40; 

the redangle AB, BC to AE, BC that is«, to the parallel©- ^ ?5. 1. 
gram AC is given. 

And it is evident how the ratio of the re£langle to the pa- 
rallelogram may be found, by making the angle FGH equal 
to the given angle ABC, and dravvmg, from any poin" F ia 
one of its sides, t K perpendicular to the other GH ; for GF 
is to FK, as BA to A£, that is, as the reitangle AB, BC, to 
the parallelogram AC. 

Cor. And if a triangle ABC has a given angle ABC, the 66. 
re<Stangle AB, BC contained by the sides about that angle, 
shall have a given ratio to the triangle ABC. 

Complete the parallelogram ABCD ; therefore, by this pro- 
position, the rectangle AB, BC has a given ratio to the paral- 
lelogram AC ; and AC has a given ratio to its half the tri- 
angle** ABC; therefore the reitangle AB,.BC has a given 4 41 1 
'ratio to the triangle ABC. , q ^^^ 

And the ratio of ihe rectangle to the triangle is found thus : 
Make the triangle FGK, as was shewn in the proposition : the 
ratio of GF to the half of the perpendicular f K is the same with 
the ratio of the redangle AB, BC to the triangle ABC. Be- 
cause, as was shewn, GF is to FK, as AB, BC to the paralle- 
logram AC ; and FK is to its half, as AC is to its half, which 
is the triangle ABC ; therefore, ex aequali, GF is to the half 
of FK, as AB, BC redangle is to the triangle ABC. 

PROP. LXIII. 56, 

X F two parallelograms be equiangular, as the side 
of the first to a side of the second, so is the other 
side of the second to the straight line to which the 
other side of the first has the same ratio which the 
first parallelogram has to the second. And conse- 
quently, if the ratio of the first parallelogram to the 
second be given, the ratio of the other side of the 
first to that straight line is given ; and if the ratio 
of the other side of the first to that straight line be 
given, the ratio of the first parallelogram to the se- 
cond is given. 

Let AC, DF be two equiangular parallelograms, as BC, a 
side of the first, is to EF, a side of the second, so is DE, the 
other side of the second, to the straight line to which AB, the 
other side of the first has the same rAtio which AChastd DF. 

D d 4 Produce 



4o8 EUCLID'S 

Produce the straight line AB, and make as BC to;EF, so 
DE to BG, and complete the parallelo- 
gram BGHC ; therefore, because BC J ^ 
or GH, is to EF, as DE to BG, the r 7 
sides aDout the equal angles BGH, DEF ^/ / 

are reciprocally proportional ; where- r^ 7'C 

a 14. 6. fore* the parallelogram BH is equal to (J^i- tr 

DF ; and AB is to BG, as the paral- j 

lelogram AC is to BH, that is, to PF; -y 7 

as therefore BC is to EF, so is DE to / / 

BG, which is the straight line to which K^ F 

A.B has the same ratio that AC has to 
DF. ^ 

And if the ratio of the parallelogram AC to DF be given, 
then the ratio of the straight line AB to BG is given ; and if 
the ratio of AB to the straight line BG be given, the ratio of 
the parallelogram AC to DF is given. 

74.73. PROP. LXIV. 

See N. Iy two parallclograms have unequal but given an- 
gles, and if as a side of the first to a side of the se- 
cond, so the other side of the second be made to a 
certain straight. hne ; if the ratio of the first paral- 
lelogram to the second be given, the ratio of the 
other side of the first to that straiglit line shall be 
given. And if the ratio of the other side of the first 
to that straight line be given, the ratio of the first 
parallelogram to the second shall be given. 

Let ABCD, EFGH be two parallelograms which have the 
unequal but given angles ABC, EFG; and as BC to FG, 
so make EF to the straight line M. If the ratio of the paral- 
lelogram AC to EG be given, the ratio of AB to M is given. 
At the point B of the straight line BC make the angle 
CBK equal to the angle EFG, and complete the parallelogram 
KBCL. And because the ratio of AC to EG is given, and that 

» 35. 1 . AC is equal* to the parallelogram KC, therefore the ratio of 
KC to EG is fr'iven ; and KC, EG are equiangular j there- 

"CS. dat. fore as BC to FG, so is** EF to the straight line to which KB 
has a given ratio, viz. the same which the parallelogram 
KC has to EG ; but as BC to FG, so is EF to the straight 
liiie M i therefore KB has a given ratio to M ; and the ratio 




DATA, 

of AB to BK is given, because the triangle ABK is given In 
species'^ ; therefore the ratio of AB to M is given"*. ^ q^^^** 

And if the ratio of AB to M be given, the ratio of the ' ** 
parallelogram AC to EG is given : for since the ratio of KB to 
BA is given, as also the ratio of AB ^ 
to M, the ratio of KB to M is given"* ; IV^A.^ 
and because the parallelograms KC, EG 
are etiuianguiar, as BC to FG, so is 
*'EF to the straight line to which KB 
has the same ratio which the parallelo- 
gram KC has to EG ; but as BC to FG, 
so is EF to Al ; therefore KB is to M, 
as the parallelogram KC is to EG ; and 
the ratio of KB to M is given, therefore the ratio of the pa- :j. 
rallelogram KC, that is, of AC to EG^ is given. 

Cor. And if two triangles ABC, EFG, have two equal 
angles, or two unequal, but given, angles ABC, EFG, and if 
as BC a side of the first to FG a side of the second, so the other 
side of the second EF be made to a straight line M; if the ratio 
of the triangles be g^iven, liie ratio of the other side of the first 
to the straight line M is given. 

Complete the parallelograms A BCD, EFGHj and because 
the ratio of the triangle ABC to the triangle EFG is given, the 
ratio of the parallelogram AC to EG is given', because the pa- « 15. 5. 
ralielograms are double^ of the triangles : and because BC is to '41. i. 
FG, as EF to M, the ratio of AB to M is given by the 63d 
dat. if the angles ABC, EFG are equal j but if they be un- 
equal, but given angles, the ratio of AB to M is given by this 
proposition. 

And if the ratio of AB to M be given, the ratio of the pa- 
rallelogram AC to EG is given by the same propositions ; and 
therefore the ratio of the triangle ABC to EFG is given. 



PROP. LXV. 



I 



F two equiano^ular parallelograms have a given 6i. 
ratio to one another, and if one side has to one side 
a given ratio ; tlie other side shall also have to the 



rrt' 



Other side a given ratio. 



O' 



Let the two equiangular parallelograms AB, CD have a 
given ratio to one another, and let the side EB have a given 
ratio to the side FD ; the other side AE has afso a given ra- 
tio to the other side CF. 

Because 



B F i) 



410 E U C L I D'S 

Because the two equiangular parallelograms AB, CD have a 
given ratio to one another: as EB, a side of the first, is to FD, 

» 63. dat. a side of the second, so is^ FC, the other side of the second, to 
the straight line to which AE, the other side of the first, has 
the same given ratio which the first parallelogram AB has 
to the other CD. Let this straight line be EGj therefore the 
ratio of AE to EG is given ; 

and EB is to FD, as FC to \ ^ ^>^ 

EG, therefore the ratio of 
FC to EG is given, because 
the ratio of EB to FD is given ; 
and because the ratio of AE 
to EG, as also the ratio of 
FC to EG is given ; the ratio ■ jFT R T 

'' 9. dat. of AE to CF is given''. 

The ratio of A E to CF may be found thus : Take a straight 
line H given in magnitude ; and because the ratio of the paral- 
lelogram AB to CD is given, make the ratio of H to K the 
same with it. And because the ratio of FD to EB is given, 
make the ratio of K to L the same : The ratio of AE to CF is 
the same with the ratio of H to L. Make as EB to FD, so FC 
to EG, therefore, by inversion, as FD to EB, so is EG to FC ; 
and as AE to EG, so is'' (the parallelogram AB to CD, and so 
is) H to K ; but as EG to FC, so is (FD to EB, and so is) K 
to L i therefore, ex aequali, as AE to FC, so is H to L. 



«f. PROP. LXVI. 

J. F two parallelograms have unequal but given an- 
gles, and a given ratio to one another; if one side 
has to one side a given ratio, the other side has also 
a given ratio to the other side. 

Let the two parallelograms A BCD, EFGH which have the 
given unequal angles ABC, EFG have a given ratio to one 
another, and let the ratio of BC to FG be given ; the ratio 
also of AB to EF is given. 

At the point B of the straight line BC make the angle CBK ; 
equal to the given angle EFG, and complete the parallelo- 
gram BKLC ; and because each of the angles BAK, AKB isj 
, 1 dat g'^^"> ^^^ triangle ABK is given* in species j therefore thej 
ratio of AB to BK is given j and because, by the hypothesis,; 

the 



DATA. 



41 « 



* 9. dat. 



the ratio of the parallelogram AC to I'.G i> given, and that AC 
is equaP to BL ; therefore the ratio of BL to EG is given - ^^S I 
and because BL is equiangular to EG, and by the hypothesis, 
the ratio of BC to FG is given ; thereforg*^ the ratio of K B to « 65. dn. 
EF is given, and the ratio of KB 
to BA is given ; the ratio there- 
fore"* of AB to EF is given. 

The ratio of AB to £F may be 
found thus ; Take the straight line 
MN given in position and magni- 
tude ; and make the angle MN O 
equal to the given angle BAK, 
and the angle MNO equal to the 
given angle EFG, or AKB : And 

because the parallelogram BL is equiangular to EG, and has a 
given ratio to it, and that the ratio of BC to FG is given ; find 
by the 65th dat. the ratio of KB to KF : and make the ratio of 
NO to (J? the same with it; Then the ratio of AB to EF is 
the same with the ratio of MO to OP : For since the tri- 
an»l« ABK is equiangular to MON, as AB to BKj so is 
Mb to ON : And as KB to EF, so is NO to OP j therefore 
exacquali, as AB to EF, so is MO to OP. 




PROP. LXVIL 



70. 



JLF the sides of two equiangular parallelograms See n. 
have given ratios to one another; tl\e parallelo- 
grams shall have a given ratio to one another. 

h Let ABCD, FFGH be twd equiangular parallelograms, and 
let the ratio of AB to FF, as also the ratio of BC to FG, be 
given ; the ratio of the parallelograrn AC to EG is given. 

Take a straight line K given in magnitude, and because the 
ratio of AB to EF is given 
make the ratio of K to L the 
same with it -, therefore L is 
given*: And because the ratio 
of BC to FG is given, make 
the ratio of L to M the same : 
Therefore M is given*, and 
K is given; wherefore** the 
ratio of K to M is given : But thb parallelogram AC is to the ^ i ^at 
parallelogram EG, as the straight line K to the straight line M, 

as 




• 2. dat. 



412 EUCLID'S 

as is demonstrated in the 23d Prop, of B. 6. Elem. therefore 
the ratio of AC to EG is given. 

From this it is plain how the ratio of two equiangular paral- 
lelograms may be found when the ratios of their sides are given. 



TO. 



PROP. LXVIII. 



SoeN. 



» 43. dat. 



«> 9. dat. 



«67,dat. 



^ 35. 1. 



XF the sides of two parallelograms which have un- 
equal, but given angles, have given ratios to one 
another ; the parallelograms shall have a given ratio 
to one another. 

Let two parallelograms ABCD, EFGH which have the 
given unequal angles ABC, EFG have the ratios of their 
sides, viz. of AB to EF, and of BC to FG, given ; the ratio 
of the parallelogram AC to EG is given. 

At the point B of the straight line BC, make the angle CBK 
equal to the given angle EFG, and complete the parallelo- 
gram KBCL : And because each of the angles BAK, BKA 
is given, the triangle ABK is given* inspecics : Therefore the 
ratio of A B to BK is given ; and the ratio of AB to EF'is 
given, wherefore^ the ratio of BK to EF is given. And the 
ratio ofBC to FG is given; r t^ 17 ft 

and the angle KBC is equal &Jk =Ll_1> KJkl 

to the angle EFG ; there- 
fore*^ the ratio of the paral- 
lelogram KC to EG is 
given : But KC is equal'^ to 
AC ; therefore the ratio of 
AC to EG is given. 

The ratio of the parallelogram AC to EG may be found 
thus : Take the straight line IVIN given in positior and mag- 
nitude, and make the angle MNO equal to the given angle K AB, 
and the angle NMO equal to the given angle AKB or FEH : 
And because the ratio of AB to EB" is given, make the ratio of 
NO toP the same ; also make the ratio of P to Q_the same 
with the given ratio of BC toFG, the parallelogram AC is to 
EG, as MO to Q; 

Because the angle KAB is equal to the angle MNO, and 
the angle AKB equal to the angle NMO j the triangle AKB 
is equiangular to NMO : Therefore as KB to B A, so is MC9 
to ON i and as BA to EF, so is NO to P j wherefore, ex 
aequall, as KB to EF, so is MO to P: And BC is to FG, as P 

3 to 







DATA. 413 

to CL) and the parallelograms KC, EG are equiangular ; there- 
fore, as was shewn in Prop. 67, the parallelogram KC, that 
is, AC is to EG, as MO to Q. 

Cor. I. If two triangles ABC, DEF have two equal an- "'- 
gles, or two unequal, but given angles ABC, DEF, and if the 
ratios of the sides about these angles, . _, 

viz. the ratios of A B to DE, and of ^ Y D H 

BC to EF be given ; the triangles 
shall have a given ratio to one ano- 
ther. 3 c g f 

Complete the parallelograms, BG, 
EH ; the ratioof BGto EH is given^ ; and therefore the tri- .ejorf-i. 
angles which are the halves'^ of them have a given' ratio to h 34''", 
one another. " '^- ^• 

Cor. 2. If the bases BC, EF of two triangles ABC, E)EF '^• 
have a given ratio to one another, and if also the straight lines 
AG, DH which are drawn to the bases from the opposite an- 
gles, either in equal angles, or unequal, but given angles AGO, 
DHF have a given ratio to one j^ j^ L D 

another; thetrianglesshall have 
a given ratio to one another. 

Draw BKjELparallel to AG, 



/ 



ti 




DH, and complete the paral-'-> Vx C jlL. Jfi F 
lelograms KC, LF. and because the angles AGC, DHF, or 
their equals, the angles KBC, LEF are either equal, or un- 
equal, but given ; and that the ratio of AG to DH, that is, of 
KB to LE, is given, as also the ratio of BCto EF ; therefore* ,67 or 68. 
the ratio of the parallelogram KC to LF is given ; wherefore '* 
also the ratio of the triangle ABC to DEF is giveIl^ "^tl 3" 



61. 



PROP. LXIX. 

i F a parallelogram which has a given angle he ap- 
plied to one side of a rectilineal figure given in spe- 
cies ; if the figure have a given ratio to the parallelo- 
gram, the parallelogram is given in species. 

Let ABCD be a rectilineal figure given in species, and to 
one side of it AB, let the parallelogram ABEF having the 
given angle ABE be applied ; if the figure ABCD hasagiven 
ratio to the parallelogram BF, the parallelogram BF is given 
in species. 

Through the point A draw AG parallel to BC, and through 
the point C draw CG parallel to AB, and produce GA, CB to 

the 



4H 

»i3. cief. 



» 53. ciat. 
« 9. Haf. 
".3,5. 1. 
« 1. 6, 



E U C L I D'S 

the points H, K ; because the angle ABC Is given*, and the 
ratioof ABto BC is given, the figure ABCD being given in 
species ; therefore, the parallelogram BG is given* in species. 
And b..'cause upon the same straight line AB the two rectilineal 
Htiures BD, BG given in species are described, the ratio of 
BD to BG is given'' ; and, by hypothesis, the ratio of 
BD to the parailelogram BF is given ; wherefore*' the ratio of 
BF, that is'', of ths'paralleloi;ram BH, to BG is given, and 
therefore^ the ratio of the straight line KB to BC is given ; 
and the ratio of BC to BA is given, wherefore the ratio of 
KB to BA is given'' : And because the angle ABC is given, 
theadJ4centangle ABKisgiven ; and the angle ABE is jriven, 
there to|e the remaining angle KBEis given. The angle EKB 
is also given, bece ise it is equal to the angle ABK ; theretore 
the triafigle. BKE is given in species, and consequently the 
ratio of EB to EK is given j and the ratio of KB to BA is giv^n 
wherefore'^ the latio or j^f. 
EB to BA is given ; and 
the angle ABE is given, 
therefore the parallelo- 
gram BF is given* in 
species. 

A parallelogram simi- 
lar to Bt may be found 
thus : Take a straight 
line LM given in position and magnitude^ and because the 
angles ABK, ABE are given, mj^ke the angle NLM equal to 
ABK, and the angle NI.O equal to ABE. And because the 
ratio of BF to BD is given, ma.ke tne ratio of LM to P the 
same with it ; and because the ratio of the figure BD to BG 
is given, find this ratio i)y the 53d dat. and make' the ratio of 
P to Q the same. Also, because the ratio of CB to BA is 
given, make the ratio of Qjo R the same ; and take LN equal 
to R ; through the point ivJ draw OM parallel to LN, and 
complete the parallelogram N LOS j then this is similar tpthe 
parallelogram BY. . ' 

Becausethe angle ABK is equal to NT>M,ah(rthe angle ABE 
to NLO, file angle KHE is equal to MLOj and the angles 
BKE, LMO are equal, because the angle ABK is equal to 
NLM; therefore the triangles l^KE, LMO are equiangu- 
lar to one another ; wherefore, as BE to BK, so is LO to 
LM ; and because as the figure BF to BD, so is the straight 
line LM to P; and as BD to BG, so is P to Qj ex aequali, 
as BF, that is**, BPI to BG, so is LM to Q ; brut EH is to^ 

BG, 




DATA. 415 

BG, as KB to BC ; as therefore KB to BC, so is LM to Qj 
and because BE is toBK as LO to LM ; and as BK to BC, 
so is LM to Q^: And as BC to BA, so Q^was made to R ; 
therefore, ex sequali, as BE to BA, so is LO to R, that is, 
to LN ; and the angles ABE, NLO are ecjual ; therefore the 
parallelogram BF is similar to LS. 

PROP. LXX. 62. 7». 

JLF two straight lines have a given ratio to one ano- See n. 
ther, and upon one of them be described a rectili- 
neal figure given in species, and upon the other a 
parallelogram having a given angle; if the figure 
have a given ratio to the parallelogram, the paral- 
lelogram is given in species. 

Let the two straight lines AB, CD have a given ratio to 
one another, and upon AB let the figure AEB given in species 
be described, and upon CD the parallelogram DF having th« 
given angle FCD ; if the ratio of AEB to DF be given, the 
parallelogram DF is given in species. ' 

Upon the straight line AB, conceive the parallelogram AG 
to be described similar, and similarly placed to FD ; and because 
the ratio of AB to CD is given, and upon them are described 
the similar rectilineal figures AG, 
FD i the ratio of AG to FD is 
given*) and the ratio of FD to AEB 
is given ; therefore '' the ratio of 
AEB to AG is given ; and the angle 
ABG is given, because it isequal to 
the angle FCD j because therefore 
the parallelogram AG which has a 
given angle ABG is applied to a 
side AB of the figure AEB given 
in species, and the ratio of AEB to AG is given, the parallelo- 
gram AG is given'' in species ; but FD is similar to AG j «: 69. dat. 
therefore FD is given in species. 

A parallelogram similar to FD may be found thus : Take 
a straight line H given in magnitude ; and because the ratio of 
the figure AEB to FD is given, make the ratio of H to K the 
same with it : Also, because the ratio of the straight line CD 
to AB is given, find by the 54th dat. the ratio which the figure 
FD described upon CD has to the figure AG described upon 
AB similar to FD j and make the ratio of K to L the same 
with this ratio : And because the ratios of H to K, and of K 

to 




4i6 EUCLID'S 

i> p. dat. to L arc given, the ratio of H to L is given^ j because, the re- 
fore, ?s AEB to FD, so is H to K : and as FD to AG, so is K 
to L i ex lequali, as AEB to AG so is H to L ; therefore the 
ratio of AEB to AG is given ; and the figure AEB is given in 
species, and to its side AB the parallelogram AG is applied in 
the given artgle ABG ; therefore by tne 69th dat. a parallelo- 
gram may be found similar to AG : Let this be the parallelo- 
gram MN ; MN also is similar to FD ; for, by the construc- 
tion, MN is similar to AG, and AG is similar to FD; there- 
fore the parallelogram FD is similar to MN. 

PROP. LXXI. 

^*- 1_F the extremes of three proportional straight 
lines have given ratios to the extremes of other 
three proportional straight hnes ; the means shall 
also have a given ratio to one another: And if one 
extreme has a given ratio to one extreme, and the 
liiean to the mean ; likewise the other extreme shall 
have to the other a given ratio. 

Let A, B, C be three proportional straight lines, and D, E, 
F, three other ; and let the ratios of A to D, and of C to F, be 
given J then the ratio of B to E is also given. 

Because the ratio of A to D, as also of C to F, is given, the 

» 67. dat. ratio of the rectangle A, C to the rectangle D, F' is given*; 

b n. 6. but the square of B is equal"" to the rectangle A, C ; and the 

square of £ to the retflangle'' D, F ; therefore the ratio of the 

«58. daf. square of B to the square of E is given ; wherefore"^ also the 

ratio of the straight line B to E is given. 

Next, let the ratio of A to D, and of B to E, be 
given ; then the ratio of C to F is also given. -' *■'*'' 

Because the ratio of B to E is given, the ratio of l-k X 

* 54. Jat. -^^g square of B to the square of E is given'' ; there- 1~~* 

fore'' the ratio of the redlangle A, C to the redangle Y 1 

D, F is given; and the ratio of the side A to the | 

side D is given ; therefore the ratio of the other ( 

• e ■. flat, sjjjg Q jQ t)^e other F is given''. 

Cor. And if the extremes of four proportionals have to the 
extremes of foijr other proportionals given ratios, and one of 
the means a given ratio to one of the means ; the other mean 
shall have a given ratio to the other mean, as may be shewn 
in the same mamier as in the foregoing proposition. 



DATA. 



417 



PROP. LXXII. 

1 F four straight lines be proportionals; as the first 
is to the straight line to which the second has a 
given ratio, so is the third to a straight line to 
which the fourth has a given ratio. 

Let A, B, C, D be four proportional straight lines, viz. as 
A to B, so C to D ; as A is to the straight line to which B has 
a given ratio, so is C to a straight line to which D has a given 
ratio. 

• Let E be the straight line to which B has a given 
ratio, and as B to E, so make D to F : The ratio of 
B to E is given*, and therefore the ratio of D to F ; 
and because as A to B, so is C to D ; and as B to E, . 

$0 D to F ; therefore, ex aequali, as A to E, so is A. x> xl 
C to F ; and E is the straight line to which B has a C D F 
given ratio, and F that to which D has a given ratio ; 
therefore as A is to the straight line to which B has 
a given ratio, so is C to a line to which D has a 
given ratio. 



S2. 



•H> 



PKOP. lxxiil 



83. 



J.F four straight lines be proportionals ; as the first seeN. 
is to the straight line to which the second has a 
given ratio, so is a straight line to which the third 
has a given ratio to the fourth. 

Let the straight line A be to B, as C to D ; as A to the 
straight line to which B has a given ratio, so is a 
straight line to which C has a given ratio to D. 

Let E be the straight line to which B has a 
given ratio, and as B to E, so make F to C ; because 
the ratio of B to E is given, the ratio of C to F is 
given : And because A is to B, as C to D ; and as 
B to E, so F to C : therefore, ex aequali, in pro- 
portLone perturbato", A is to E, as F to D ; that is, | • J3. 5. 

A is to E to which B has a given ratio, as F, to 
which C has a given ratio, is to D. 

Ee 



A BE 

FCD 



4i8 E U C L I D'S 

Si. PROP. LXXIV. 

J F a triangle has a given obtuse angle ; the excess 
ot" the square ot the side which subtends the obtuse 
angle, above the squares of the sides which contain 
it, shall have a given ratio to the triangle. 

Let the triangle ABC have a given obtuse angle ABC j and 
produce the siraight line CB, and £rom the point A draw AD 
perpendicular to BC : The excess of the square of AC above 

».12 2. the squares of AB, BC, that is% the double of the rectangle 

contained by DB^ BC, has a given ratio to the triangle ABC. 

Because theangle ABCis given, theangle A BD is also given ; 

and the angle ADB is given ; wherefore the triangle ABD 

"43. «iat. is given'^ in species ; and therefore the ratio of AD to DB 

M. 6. is given : And as AD to DB, so is<^ the re£langle AD, BC 
to the rectangle DB, BC ; whercFore th» ratio of the re6^- 
angle AD, BC to tKe reiiangle DB, BC is given, as also the 
ratio of twice the rectan^j^le DB, BC to 
the re6langle AD, BC : But the ratio of 
the i entangle A D, BC to the triangle ABC 

"41.1. is given, because it is double** of the tri- 
angle ; therefore the ratio of twice the 
redtangle DB, BC to the triangle ARC is 

"^^.dat. given"^; and twice the redlangle-DB, BC 

IS the excess^ of the square of AC above the squares of AB, 
BC J therefore this excess has a given ratio to the triangle 
ABC. 

And the ratio of this excess to the triangle ABC naay be 
found thus: Take a straight line EF given in position and mag- 
nitude i and because the angle ABC is given, at the point F 
of the strai:;ht line F,F, make the angle EFG equal to the 
ang|e ABCj produce GF, and draw EH perpendicular to 
FG; then the ratio of the excess of the square of AC above 
the squares of AB, BC to the triangle ABC, is the same with 
the ratio of quadruple the straight line HF to HE. 

Because the angle ABD is equal to the angle EFH, and 
the angle ADB to EHF, each being a right angle : tjje tri- 

'4.6. angle ADB is equiangular to EHF; therefore*" as BD to DA, 

«Cor. 4.5. so FH to HE; and as quadruple of BD to DA, so is 8 qua- 
druple of B'H to HE : Biit as twice BD is to DA, so is*^ twice 
the redangle DB, BC o the redangle AD, BC ; and as DA 

T. 5. to the half of it, so i.s'' the rectangle AD, BC to its half the 

triangle 




C5. 



DATA. 

triangle ABC ; therefore, ex aequali, as twice BD is to the half 
of DA, that is, as quadruple of BD is to D.A, that is, as qua- 
druple of FH to HE, so is twice the reJlaiigle DB, BC to the 
triangle ABC. 

PROP. LXXV. 

1 F a triangle has a given acute angle, the space by 
which the square of the side subtending the acute 
angle is less than the squares of the sides which con- 
tain it, shall have a given ratio to the triangle. 

Let the triangle ABC have a given acute angle ABC, and 
draw AD perpendicular to BC, the space by which the square 
of AC is less than the squares of AB, BC, that is% the double » 13. 2. 
of the rectangle contained by CB, BD, has a given ratio to the 
triangle ABC. 

Because the angles ABD, ADB are each of them given, 
the triaagle ABD is given in species ; and therefore the ratio 
of BD to DA is given : And as BD to DA, 
so is the re<5tangle CB, BD to the re<Ebng!e 
CB, AD : Therefore the ratio of these re<£l- 
ang!cs is given, as also the ratio of twice the 
reaangle CB, BD, to the reclangle CB, AD, 
but the rectangle CB, AD has a given ratio 
to its half the triangle ABC : Therefore'' the 
ratio of twice the redanglc CB, BD to the triangle ABC is 
given : and twice the rectangle CB, BD is* the space by which 
the square of AC is less than the squares of AB, BC ; there- 
fore the ratio of this space to the triangle ABC is given : And 
the ratio may be found as in the preceding proposition. 

LEMMA. 

I F from the vertex A of an isosceles triangle A BC, any straight 
line AD be drawn to the base BC, the square of the side 
AB is equal to the rectangle BD, DC of the segments of the 
base together with the square of AD ; but if AD be drawn to 
the base produced, the square of AD is equal to the rectangle 
BD, DC, together with the square of AB. 
Case i. Bisect the base BC in E, and 
join AE, which will be perpendicular* to 
BC J wherefore the square of AB is equal 
*• to the squares of AE, EB; but the square 
of EB is equal* to the re«angle BD, DC 
together with the square of DE ; there- 
fore the square of AB is equal to the 

E e 2 squares 



419 




DC 



9. dat 




»8. 1. 



^47. 1. 



\2<5 E U C t I D ' S 

MT. ;. squares of AE, fcD, that is, to^ the square of AD, together 
with the rectangle BD, DCj the other case is shewn in the 
same way by 6. 2. Elem. 

ej. PROP. LXXVl. 

xF a triangle have a given angle, the excess of the 
square of the straight line which is equal to the two 
sides that contain the given angle, above the square of 
the third side, shall have a given ratio to the triangle. 

/ Let the triangle ABC have the given angle BAC, the excess 
of the square oT the straight line which is equal to BA, AC 
together above the square of BC>.-shalI have a given ratio to 
the triangle ABC. 

Produce BA, and take AD equal to AC, ioin DC, and 
produce it to E, and through the point B draw BE parallel to 
AC; join AE, and draw AF perpendicular to DC ; and be- 
cause AD is equal to AC, BD is equal to BE ; and BC is 
drawn from the vertex B of the isosceles triangle DBE ; there- 
fore, by the Lemma, the square of BD, that is, of BA and 
AC together, is equal to the rectangle DC, CE together with 
the square of BC ; and therefore the square of BA, AC to- 
gether, that is, of BD, is greater than 
the square of BC by the rectangle DC, 
CE ; and this rectangle has a given 
ratio to the triangle ABC, because 
the angle BAC is given, the adjacent 
angle CAD is given ; and each of the 
, angles ADC, DCA is given, for 

•5. &: 32. each of them is the half^ of the given ^^^^^^_\ 

'• angle BAC; therefore the triangle ^ \~!L 

*43. dat.' ADC is givent> in species ; and AF is EL 

drawn from its vertex to the base in 
■^ a given angle ; wherefoie the ratio of AF to the base CD is 
* 50. dat. given" and as CD to AF, so is"^ the rectangle DC, CE to 
" 1- <3- the rectangle AF, CE ; and the ratio of the rectangle AF, 
'41.1. CE to its hall"=, the triangle ACK, is given; therefore the 
' '". I. ratio of the rectangle DC, CE to the triangle ACE, that is', 
« 9, dat, to the triangle ABC, is given? ; and the rectangle DC, CE 
is the excess of the square of BA, AC, together above the 
square of BC : Therefore the ratio of this excess to the tri- 
angle ABC is given. 

The ratio which the rectangle DC, CE, has to the triangle 
ABC is found thus : Take the straight line HG given in posi- 
tion 




6. 
no 5. 



DATA. 4H 

tion and magnitude, and at the point G in GH make the angle 
HGK equal to the given angle CAX>, and take GK equal to 
GH, join KH, and draw GL perpendicular to it : Then the 
ratio of HK to the half of GL is the same with the ratio of 
the rectangle DC,CE to the triangle ABC : Beouse the angles 
HGK, DAC, at the vertices of the isosceles triangles GHK, 
ADC, are equal to one another, these triangles are similar i 
and because GL, AF, are perpendicular to the bases HK,DC, 
as HK to GL, so is ^ (DC to AF, and so is) the rectangle "> i^ 
DC, C£ to the rectangle AF, CEj but as GL to its half, so 
is the rectangle AF, CE to its half, which is the triangle ACE, 
or the triangle ABC;- therefore, ex acquaii, HK is to the half 
of the straight Line GL, as the rectangle DC, C£, ir to the 
triangle ABC. - •- 

Cor. And if a triangle have a given angle, the spacE by 
which the square of the straight line, which is the difference 
of the sides which contain the given angle, is less than the 
square of the third side, shall have a given ratio to the trian- 
gle. This is demonstrated the same way as in the preceding 
proposition, by help of the second case of the Lemma. 

PROP. LXXVIL I. 

If the perpendicular drawn from a given angle of as*eN. 
triangle to the opposite side, or base, has a given 
ratio to the base, the triangle is given in species. 

Let the triangle ABC have the given angle BAC, and let the 
perpendicular AD drawn to the base BC, have a given ratio 
to it, the triangle ABC is given in species. 

If ABC be an isosceles triangle, it is evident*, that if any 



5.1c 22. 
1. 




A 



B Pv D C 




one of its angles be given, the rest are also given ; and there- 
fore the triangle is given in species, without the consideration 
©f the ratio of the perpendicular to the base, which in this case 
is given by Prop. 50. 

But when .-VBC is not an isosceles triangle, take any straight 
line EF given in position and magnitude, and upon it descr-.be 

E e 3 the 



422 



E U C LI D'S 



the segment of a circle EGF, containing an angle equal to thf 
given angle BAC, draw GH bisecting EF at right angles, and 
join EG, OF : Then, since the angle EGF is equal to the angle 
BAC, and that EGF is an isosceles triangle, and ABC is not 
the angle FEG is not ifequal to the angle CBA : Draw EL 
making the angle FEL equal to the angle CBA ; join FL, and 
drawLM perpendicular to EF; then because the triangles ELF 
BAC are equiangular, as also are the triangles MLE, DAB, 
as ML to LE, so is DA to AB ; and as LE to EF, so is A B to 
BC; wherefore, ex aequali, as LM to EF, so is AD to BC ; 
and because the ratio of AD to BC is given, therefore the ratio 
of LM to EF is given ; and EF is given, wherefore^ LM also 
is given. Complete the parallelogram LMFK ; and because LM 
is given, FK is given in magnitude; it is also given in position; 
and the point F is given, and consequently*^ the point K ; and 
because through K the straight line KL is drawn parallel to EF, 
" 3J. da', which is given in position, therefore** KL is given in position : 



2. dat. 



« 30. da' 



<= 28. dat. 
♦ -19. dat. 
« 42. dat. 





K D C 



and the circumference ELF is given in position ; therefore the 
point L is given^. And because the points, L, E, F, are given, 
the straight lines LE, EF\ FL,are given*^in magnitude ; there- 
fore the triangle LEF is given in species 8 ; and the triangle 
ABC is similar to LEF, whereforealso ABC isgiven inspecies. 

Because LM is less than GH, the ratio of LM to EF, that 
is, the given ratio of AD to BC, must be less than the ratio of 
GH to EF, which the straight line, in a segment of a circle 
containing an angle equal to the given angle, that bisects the 
base of the segment at right angles, has unto the base. 

Cor. I. If two triangks, ABC, LEF have one angle BAC 
equal to one angle ELF, and if the perpendicular AD be to the 
base BC, as ihe perpendicular LM to the base EF, the trian- 
gles ABC, LEF are similar. 

Describe the circle EGF about the triangle ELF, and draw 
LN parallel to EF, join EN, NF, and draw NO perpendicu- 
lar to EF i because the angles ENF, ELF are equal, and that 

the 



DATA. 423 

the> angle EFN is equal to the alternate angle FNL, that is, 
to the angle FEL in the same segment ; therefore the triangle 
N£F is similar to LEF; and in the segment EGF there can 
be no other triangle upon the base KF, which has the ratio of 
its perpendicular to that base the same w th the ratio of LM 
or NO to EF, because the perpendicular must be greater or 
less than LM or NO ; but, as has been shewn in the preceding 
demonstration, a triangle, similar to ABG, can be described in 
the segment EGF upon the base EF, and the ratio of its per- 
pendicular to the base is the same, as was there shewn, with 
the ratio of AD to BC, that is, of LM to EF ; therefore that 
triangle must be either LEF, or NEF, which therefore are 
similar to ihe triangle ABC. 

Cor. 2. If a triangle ABC has a given angle BAC, and if 
the straight line AR drawn from the given angle to the op- 
posite side BC, in a given angle ARC, has a given ratio to 
BC, the triangle ABC is given in species. » 

Draw AD perpendicular to BC ; therefore the trianj^lc 
ARD is given in species j wherefore the ratio of AD to AS. is 
given : and the ratio of AR to BC is given, and consequently'' " 9. dat. 
the ratio of AD to BC is given; and the triangle ABC is 
therefore given in species'. '77.dat. 

Cor. 3. If two triangles ABC, LEF have one angle BAC 
equal to one angle ELF, an ! if straight lines drawn from 
these angles to the bases, making with chem given and equal 
angles, have the same ratio to the bases, each to each ; then 
the triangles are similar; for having drawn perpendiculars t« 
the bases from the equal angles, as one perpendicular is to its 
base, so is the other to its base^j wherefore, by Cor. i. the "5 22. 5. 
triangles are similar, 

A triangle similar to ABC may be found thus : Having de- 
scribed the segment EGF, and drawn the straight line GH as 
was dlre£led m the j^oposition, find FK, which has to EF the 
given ratio of AD to BC ; and place FK at right angles to EF 
from the point F ; then because, as has been >hewn, the ratio 
of AD to BC, that is, of FK to EF, must be less than the ratio 
of GB to EF ; therefore FK is less than G H ; and consequently 
the parallel to EF drawn through the point K, must meet the 
circumference or the segment in two points: Let L be either 
of them, and join EL, LF, and draw LM perpendicular to 
EF : then, because the angle BAC is equal to the anijle 
ELF, and that AD ts^to BC, as KF ; that is, LM to EF, The 
triangle ABC is similar to the triangle LEF, by Cor. i. 

E 64 



424 



fe U C L I D'S 



80. 



'41. 1. 

' Cor. 62. 
dat. 

9. dat. 
'1.6. 
77. dat. 



PROP. LXXVIII. 

IF a triangle have one angle given, and if the ra- 
tio of the rectangle of the sides which contain the 
given angle to the square of the third side be given, 
the triangle is given in species. 

Let the triangle ABC have the given angle BAC, and let 
the ratio of the reftangle BA, AC to the square of BC be 
given J the triangle ABC is given in species. 

From the point A, draw AD perpendicular to BC, the re6^- 
angle AD, BC has a given ratio to its half* the triangle ABC j 
and becausfe the angle BAC is given, the ratio of the triangle 
BAC to the rectangle B A, AC is given^ j and by the hypo- 
thesis, the ratio of the redlangle BA, AC to the square of BC is 
given j therefore' the ratio of the redangle AD, BC to the 
square of BC, that is*^, the ratio of the straight line AD to BC 
is given ; wherefore the triangle ABC is given in species'^. 

A triangle similar to ABC may be found thus : Take a 
straight line EF given in position and magnitude, and make 
the angle FEG equal to the given angle BAC, and driaw FH 
perpendicular to EG, and BK perpendicular to AC : therefore 
thetrianglesABKjEFH 
are similar, and the reft- a 
an2;le AD, BC or the 
reaangle BK,AC which 
is equal to it, is to the 
rectangle B A, AC as the 
straight line BK to BA, ii K 
thatis,asFHtoFE. Let 
the given ratio of the rectangle B A, AC to the square of BC 
be the same with the ratio of the straight line EF to FL j there- 
fore, ex aequali, the rjrtio of the reftangle AD, BC to the 
square of BC, that is, the ratio of the straight line AD to BC, 
is the same with the ratio of HF to FL ; and because AD is 
not greater than the straight line MN in the segment of thfe 
circle described about the triangle ABC, whicli bisects BC 
at right angles ; the ratio of AD to BC, that is, of HF 
to FL, must not be greater than the ratio of MN to BC : 
Let ic be so; and, by the 77th dat. find a triangle OPQ 
which has one of its angles POQ^ equal to the given 
angle BAC, and the ratio of the perpendicular OR, 
drawn from that angle to the base PQ, the same with the 
ratio of HF to FL ; then the triangle ABC is similar to 

OPQ 





10. dat 



DATA. 4^5 

OPO : Because, as has been shewn, the ratio of AD to BC is 
the "same with the ratio of (HF to FL, that is, by the con- 
struction, with the ratio of) OR to PQ; ajid the angle BAG 
is equal to the angle POQ. Therefore the triangle ABC is 
similar^ to the trianele POQ. *^i. Cot. 

OthtrxeisCy 

Let tht triangle ABC have the given angle BAC, and let 
the ratio of the rectangle BA, AC tto the square of BC be 
giv^en i the triangle ABC is given in species. 

Because the angle BAC is given, the excess of the square 
of both the sides BA, AC together above the square of the 
third side BC has a given* ratio to the triangle ABC. Let the »76. dat. 
figure D be equal to this excess ; therefore the ratio of D to 
the triangle ABC is given ; and the ratio of the triangle ABC 
to the rectangle BA, AC is sriven\ because BAC is a siven „ 

angle ; and the redangle BA, AC has ^ ] ] ^^^^' 

a given ratio to the square of BC; 
wherefore*^ the ratio of D to the 
square of BC is given ; and, by cotn- 

position<*, the ratio of the space D, i^ ^ «7 dat. 

together with the square of BC to the square of BC is given; 
but D together with the square of BC is equal to the square of 
both BA and AC toother ; therefore the ratio ofthe square of 
BA, AC together to the square of BC is given ; and the ratio 
of B A, AC together to BC is therefore given'' ; and the angle 
BAC is given, wherefore^ the triangle ABC is given in species. ' ^^' ^*- 
The composition of this, which depends upon those of the ^^' ^^^ 
76th and 48th propositions, is more complex than the preced- 
ing composition, which depends upon that of Prop. 77. which 
is easy. 

PROP. LXXIX. 

K. 

IF a triangle ])ave a given angle, and if the straight see x. 
hne drawn from that angle to the base, making a 
given angle with it, divides the base into segments 
which have a given ratio to one another ; the trian- 
gle is given in species. . 

• Let the triai>gle ABC have the given angle BAC, and let 
the straight line AD drawn to the base BC, making the given 
angle ADB, divide CB in;? the segments BD, DC which hkve 



'47.dat, 




4i6 E U C L 1 D'S 

a given ratio to one another ; the triangle ABC is given in 
species 

» 5. 4. Describe* the circle BAG about the triangle, and from its 

centre E, draw EA, EB, EC, ED ; because the angle BAG is 

«» 20^3. given, the angle BEC at the centre, which is the double'' of it, 
is given. And the ratio of BE to EC is given, because they 

*4i. dat. are equal to one another; therefore the<= triangle BEC is 
given in species, and the ratio of EB to BC is given ; also the 

« 7.dat. ratio of CB to BD is given"*, because the ratio of BD to DC 

«9, dat. is given ; therefore the ratio of EB to BD is givcn%*and the 
angle EBC is given, wherefore the triangle EBD is given* 
in species, and the ratio of EB, that is, of EA, to ED, is there- 
fore cfiven ; and the angle EDA is given, because each of the 
angles BDE, BDA is given ; therefore the triangle AKD is 
given'' in species, and the angle AED 
given ; also the angle DEC is given, be- 
cause each of the angles BED, BEC is 
given ; therefore the angle AEC is given, 
and the ratio of EA to EC, which are 
equal, is given ; and the triangle AKC is 1^\ 
therefore given^ in species, and the angle 
ECA is given ; and the angle ECB is given, 
wherefore the angle ACB is given, and the angle BAC is also 
given i therefore? the triangle ABC is given in species. 

A triangle similar to ABC may be found, by taking a straight 
line given in position and magnitude, and dividing it in the 
given ratio which the segments BD, DC are required to have 
to one another ; then, if upon that straight line a segment of a 
circle be described containing an angle equal to the given angle 
BAC, and a straight line be drawn from the point of division 
in an angle equal to the given angle ADB, and from the point 
where it meets the circumference, straight lines be drawn to the 
■extremity' of the first line, these, together with the first line, 
shall contain a triangle similar to ABC, as may easily be shewn. 
The demonstration may be also made in the manner of that 
of the 77th Prop, and that of the 77th may be made in the 
manner of this. 

PROP. LXXX. 

XF the sides about an angle of a triangle have a gi- 
ven ratio to one auotlier, and if the perpendicular 
drawn from tliat angle to the base has a given ra- 
tio to the base ; the triangle i; given in species. 

Lst 



« 43.*dat. 



DATA. 427 

Let the sides BA, AC, aboQt the angle BAG of the trian- 
gle ABC/ have a given ratio to one another, and let the per- 
pendicular AD have a given ratio to the base BC, the triangle 
ABC is given in species. 

First, let the sides AB, AC be equal to one another, there- 
fore the perpendicular AD bisects- the base *26. 1, 
BC; and the ratio of AD to BC, and there- ^ 
fore to its halt DBy is given ; and trie angle < / \ ..^ 
ADB is given; wherefore the triangle* ABD y. \ •'^*'- 
and consequently the triangle ABC is given'' p rv ~p '144.dat, 
in species. 

But let the sides be unequal, and BA be greater than AC ; 
and malce the angle CAE equal to the angle ABC ; because 
the angle AEB is common to the triangleb AEB, CEA, they 
are similar ; therefore as AB to BE, so is CA to AE, and, by 
permutation, as BA to AC, so is BE to EA, and so is EA 
to EC ; and the ratio of BA to AC is given, therefore the 
ratio of BE to E A, and the ratio of EA to EC, as also the 
ratio of BE to EC is given*"; wherefore the ratio of EB tOcp ,jat- 
BCis given"*; and the ratio of AD to BC "6 dzt, 

is given by the hypothesis, therefore'^ the -^ 

ratio of AD to BE is given ; and the ratio -^ • 
of BE to EA was shewn to be given ; 
whereforetheratioof AD toEA is given; ^ [," ^ Y, E' 
and ADE is a right angle, therefore the 
triangle ADE is given' in species, and the angle AEB given; 
the ratio of BE to EA is likewise given, therefore*" the trian- *'^- *■'**• 
gle ABE is given in species, and consequently the angle E AB, 
as also the angle ABE, that is, the angle CAE is given; 
therefore the angle BAG is given, and the angle ABC being 
also given, the triangle ABC is given* in species. '^43. dat. 

How to find a triangle which shall have the things which 
are mentioned to be given in the proposition, is evident in 
the first case ; and to find it the more easily in the other 
case, it is to be observ ed that, if the straight line EF equal to 
EA be placed in EB towards B, the point F divides the base 
BC into the segments BF, FC, which have to one another the 
ratio of the sides BA, CA, because BE, EA, or EF, and •19. 5. 
EC, were shown to be proportionals, therefore* BF is to FC, 
as BE to EF, or EA, that is, as BA to AC ; and AE cannot 
be less than the altitude of the triangle ABC, but it may be 

equal 




4«8 E U C L I D'S 

equal to it, which, if it be, the triangle, in this case, as also 
the ratio of the sides, may be thus found : Having given the 
ratio of the perpendicular to the base, take the straight line 
• GH, given in position and magnitude, for the base of the tri- 
angle to be found; and let the given ratio of the perpendicular 
to the base be that of the straight line K to GH, that is, let K 
be equal to the perpendicular ; and suppose GLH to be the 
triangle which is to be found, therefore having made the angle 
HLM equal to LGH, it is required that LM be perpen- 
dicular to GM, and equal to K; and because GM, ML, 
I MH are proportionals, as was shewn of BE, EA, EC, the 

re<9angle GMH is equal to the square of ML. Add the com- 
mon square of NH (having bisected GH in N) and the square 

«6.2. of NM is equal* to the squares of the given straight lines NH 
and ML, or K i therefore the square of NM, and its side 
NM is given, as also the point M, viz. by taking the straight 
line NM, the square of which is equal to the squares of NH, 
ML. Draw ML equal to K, at right angles to GM ; and be- 
cause ML is given in position and magnitude, therefore the 
point L is given, join LG, LH ; then the triangle LGH is 
that which was to be found ; for the square of NM is equal to 
the squares of NH and ML, and takmg away the common 
square of NH, the re6l- 
angle GMH is equale to 
the square of ML ; there- 
fore as GxM to ML, so is 
ML to MH, and the tri- 

*€.6. angle LGM is'' therefore 

equiangular to HLM, and 

the angle HLM equal to Q 
the angle LGM, and the 
straight line LM, drawn from the vertex of the triangle mak- 
ing the angle HLM equal to LGH, is perpendicular to the 
base and equal to the given straight line K, as was required ; 
and the ratio of the sides GL, LH, is the same with the ratio of 
GM to ML, tKat is, with the ratio of the straight line which 
is made lip of GN, the half of the given base and of NM, the 
sqiiare of which is equal to the squares of GN and K, to the 
straight ^lin€ JC. 

And whether this ratio of GM to ML is greater or )ess 
than the ratio of the sides of any other triangle upon the basft 
GH, and of y^hich the altitude is equal to the straight line K, 

3 that 




NilR KF 



DATA. ._ 

429 

that is, the vertex of which is in the parallel to GH drawn 
through the point L, may be thus found. Let OGH be any 
such triangle, and draw OP, making the angle HOP equal to 
the angle OGH ; therefore, as before, GP, PO, PH are pro- 
portiona,ls, and PO cannot be equal to LM, because the rect- 
angle GPH would be equal to the rectangle GMH, which 
is impossible; for the point P cannot fall upon M, because O 
would then fall on L ; nor can PO be less than LM, therefore 
it is greater ; and consequently the rectangle GPH is greater 
tlan the rectangle GMH, and the straight line GP greater 
than GM : Therefore the ratio of GM to MH is greater than 
the ratio of GP to PH, and the ratio of the square of GM 
to the square of ML is therefore' greater than the ratio of' 2. Cct. 
the square of GP to the square of PO, and the ratio of the "^ ^' 
straight line GM to ML greater than the ratio of GP to PO. 
But as GM to ML, so is GL to LH ; and as GP to PO, so 
is GO to OH ; therefore the ratio of GL to LH is greater 
than the ratio of GO to OH ; wherefore the ratio of GL to 
LH is the greatest of all others; and consequently the given 
ratio of the greater side to the less must not be greater than 
this ratio. 

But if the ratio of the sides be not the same with this greatest 
ratio of GM to ML, it must necessarily be less than it : Let 
any less ratio be given, and the same thmgs being supposed, 
viz. that GH is the base, and K equal to tne altitude of the 
triangle, it may be found as follows: Divide GH in the point 
Q, so that the ratio of GQ to QH may be the same with the 
^iven ratio of the sides ; and as GQ to QH, so make GP to 
f Q and so wilK PQ b?" to PH ; wherefore the square of GP 
is to the square of FQ, as« the straight line GP to PH : ♦ 19. 5. 
And because GM, ML, MH are proportionals, the square of 
GM is to the square of ML, as' the straight line GM to MH : 
But the ratio of GQ_to Q.H, that is, the ratio of GP to PQ, 
is Ie?s than the ratio of GM to ML ; and therefore the ratio 
of the square of GP to the square of PQ^is less than the ratio 
of the square of GM to that of ML; and consequently the 
ratio of the straight line GP to PH is less than the ratio of 
GM to MH ; and, by division, the ratio of GH to HP is less 
than that of GH to HM ; wherefore'' the straight line HP is k 10. 5, 
greater than HM, and the reftangie GPH, that is, the square 
of PQ, greater than the reiftangle GMH, that is, than the 
square of ML, and the straight line PQ^is therefore greater 

than 



43° 



E U C L I D'S 

thaa M!^. Draw LR parallel to GP, and from P draw PR zt 
right angles to GP. Because PQ_is greater than ML, or PR, 
the circle described from the centre P, at the distance PQ, 
must necessarily cut LR in two points ; let these be O 8, and 
join OG, OH ; SG, SH : each of the triangles OGH, ^GH 
have the things mentioned to be given in the proposition : 
Join OP, SP J and because as GP to PQ, or PO, so is PO 
to PH, the triangle OGP is equiangular to HOP ; as, there- 
fore, OG to GP, so is HO to OP ; and, by permutation, as 
GO to OH, so is GP to PO, or PQ_; and so is GQ.to QH ; 
Therefore the triangle OGH has the ratio of its sides GO, OH, 
the same, with the given ratio of GCi to QH : and the perpen- 
dicular has to the base the given ratio of K to GH, because the 
perpendicular is equal to LM, or K : The like may be shewn 
m the same way ot the triangle SGH. 

This construction by which the triangle OGH is found, is 
shorter than that which would be deduced from the demon- 
stration of the datum, by reason that the base GH is given 
in position and magnitude, which was not supposed in the de- 
monstration : The same thing is to be observed in the next 
proposition. 

M. PROP. LXXXI. 

XF the sides about an angle of a triangle be une- 
qual, and have a given ratio to one another, and if 
the perpendicular from timt angle to the base di- 
vides it into segments that have a given ratio' to 
one another, the triangle is given in species. 

Let ABC be a triangle, the sides of which about the angle 
BAC are unequal, and have a given ratio to one another, and 
let the perpendicular AD to the base BC divide it into the 
segments BD, DC, which have a given ratio to one another, 
the triangle ABC is given in species. 

.'" Jjct, AB be greater than AC, and make the angle CAE 

cqtial to the angle ABC ; and because the angle AEB is com- 

»4. 6. mon to the triangles ABE, CAE, they are* equiangular to 

ojie another ; Therefore as AB to BE, so is CA to AE, and, 

by 




XLH ^' 



DATA. 

by permutation, as AB to AC, so BE to 
E A, and so is EA to EC : But the ratio of 
BA to AC is given, therefore the ratio 
of BE to EA, as also the ratio of EA to 
EC is given ; wherefore'' the ratio of 
BE to EC, as also"^ the ratio of EC to 
CB is given : And the ratio of BC to CD 
is given% because the ratio of BD to 
DC is given ^ therefore* the ratio of EC 
to CD is given, and consequently* the {> 
ratio of DE to EC : And the ratio of EC 
to EA was shewn to be given, therefore'' the ratio of DE to 
EA is given : And ADE is a right angle, wherefore^ the tri- 
angle A.DE is given in species, and the angle A ED given ; 
And the ratio of CE to EA is given, therefore^ the triangle 
AEC is given in species, and consequently the angle ACE is 
given, as also the adjacentangle ACB. In the same manner, 
because the ratio of BE to EA is given, the triangle BE A is 
given in species, a.d the angle ABE is theretore given : And 
the angle ACB is given ; wherefore the triangle ABC is 
givens in species. 

But the ratio of the greater side BA to the other AC must 
be less than the ratio of the greater segment BD to DC : Be- 
cause the square of BA is to the square of AC, as the squares 
of BD, DA to the squares of DC, DA ; and the squares of 
BD, DA have to the squares of DC, DA, a less ratio than the 
square of BD has to the square of DCf, because the square of 
BU is greater than the square of DC j therefore the square of 
jliA l^s to the square of AC a less ratio than the square of 
BD has to that of DC : And consequently thi ratio of BA 
to AC is less than the ratio of BD to DC. 

This being premised, a triangle w^hich shall have the things 
mentioned to be given in the projX)sition, and to which the 
triangle ABC iy similar, may be found thus : Talce a straight 
line GH given in position and magnitude, and divide it in K, 
so that the ratio ot GK to KH may be the same with the given 
ratio of BA to AC : Divide also GH in L, so that the ratio 

of 



♦31 



* 46. dat. 
'U. dat. 



« 43. dat. 



■}■ K 'i. ';? g-^ilf- (hau B, ir.l O zny :.^.-d magnitude; then A and C together 
hare to B and C tfjg- ther a l?-! ralio than A has to B. 

L-: A be to B as C to D, aud because A i* greater than B, C is greitT than I>: 
But as A i? !o P, so A and C to B and D ; and A and C have to B ani C a less ra- 
tio than A a-'d C Hnve to B and P, becau* C is greater than D, thereiore A and C 
h&«e lb 2 *pdC iea ;.itic '.hen A to B. 



y 



432 EUCLID'S 

of GL to LH may be the same with the given ratio of BD t9 
DC, and draw LM at right angles to GH : And because the 
ratio of the sides of a triangle is less than the ratio of the seg- 
ments of the base, as has been shewn, the ratio of GK tq KH 
is less than the ratio of GL to LH j wherefore the point L 
must fall betwixt K and H : Also make as GK to KH, so GN 
fciP. 5. toNK, and so shall'^NK be toNH. And from the centre N, 
at the distance NK, describe a circle, and let its circumference 
meet LM in O, and join OG, OH j then OGH is the triangle 
which was to be described; Because GN is to NK, or NO, as 
NO to N H, the triangle OGN is equiangular to HON j there- 
fore as OG to GN, so is HO to ON, and, by permutation, 
as GO to OH, so is GN to NO, or NK, that is, as GK to 
KH, that is, in the given ratio of the sides, and by the con- 
struction, GL, LH have to one another the given ratio of 
the segments of the base. 



60. PROP. LXXXIL 



XF a parallelogram given in species and magnitude 
be increased or diminished by a gnomon given in 
magnitude, the sides of the gnomon are given in 
magnitude. 

First, I>et the parallelogram AB given in species and mag- 
nitude be increased by the given gnomon ECBDFG, each of 
the straight lines CE, DF is given. 

Because AB is given in species and magnitude, and that the 
gnomon ECBDFG is given, therefore the whole space AG 
is given in magnitude : But AG is also given in species, be- 
nd cause it is similar ' to AB ; therefore the sides of AG arc 
^ j2^. 6. given^ : Each of the straight lines AE, AF 
' ^*" is therefore given j and each of the straight yL 
lines CA, AD is given'', therefore each of 
» 4.dat. the remainders EC, DF is given^ 

Next, let the parallelogram AG given in 
species, and niagnitude, be diminished by the 
given gnomon ECBDFG, each of the straight 
lines CE, DF is given. 

Because the paralleloj^ram AG is given, as 






«Jcf. 




H 



also its gnomon ECBDFG, the remaining space AB is given in 



magnitude 



DATA. 433 

magnitude : But it is also given in species ; because it is similar' » f |" ^ 
to AG ; therefore'' its sides CA, AD are given, and each of \ -ii. 6. 
the straight lines EA, AF is given ; therefore EC, DF are "'^^■d*'- 
each of them given. 

The gnomon and its sides CE, DF may be found thus in 
the first case. Let H be the given space to which the gnomon 
must be made equal, and find"* a parallelogram similar to AB "25.6. 
and equal to the figures AB and H together, and place its sides 
AE, AF from the point A, upon the straight lines AC, AD, 
and complete the parallelogram AG which is about the same 
diameter* with AB ; because therefore AG is equal to both *26. 6. 
AB and H, take away the common part AB, the remaining 
gnomon ECBDr G is equal to the remaining figure H ; 
therefore a gnomon equal to H, and its Sides CE, DF are 
found : And in like manner they may be found in the other 
case, in which the given figure t^ must be less than the figure 
FE from which it is to be taken. 



PROP. LXXXIII. 38, 

J. F a parallelogram equal to a given space be applied 
to a given straight line, deficient by a parallelogram 
given in species; the sides of the defect are given. 

Let the parallelogram AC equal to a given space be applied 
to the given straight line AB, deficient by the parallelogram 
BDCL given in species, each of the straight lines CD, DB 
are given. 

Biseft AB in E ; therefore EB is given in'magnitude ; upon 
EB describe* the parallelogram EF similar to DL and simi- * ^8. 6, 
larly placed ; therefore EF is given in 
species, and is about the same diameter'' 
with DL ; let BCG be the diameter, and 
constru£l the figure ; therefore, because 
the figure EF given in species is describ- 
ed upon the given straight line EB, EF 
is given= in magnitude, and the gno- A E D B « 56.dat. 
mon ELH is equal"* to the given figure "3^. and 

AC i therefore^ since EF is diminished by the given gnomon e 8«.'d»t. 
ELH, the sides EK, FH of the 2;nomon are given ; but EK is 
equal to DC, and FH to BD f wherefwc CD, DB are each 
of them given, 

F f TJus 




434 ' EUCLID'S 

This demonstration is the analysis of the problem in the 
28th Prop of Book 6. the construdiion and demonstration of 
which proposition is the composition of the analysis ; and be- 
cause the given space AC, or its equal the gnomon ELH, is to 
be taken from the figure EF described upon the half of AB 
similar to BC, therefore AC must not be greater than EF, as 
is shewn in the 27th Prop. B. 6. 



59. PROP. LXXXIV. 

X F a parallelogram equal to a given space be ap- 
plied to a given straight line, exceeding by a pa- 
rallelogram given in species ; the sides of the ex- 
cess are given. 

Let the parallelogram AC equal to a given space be applied 
to the given straight line AB, exceeding by the parallelogram 
BDCL given in species ; each of the straight lines CD, DB 
are given. 

Bisedl AB in E ; therefore EB is given in magnitude: Upon 

•48. 6. BE describe* the parallelogram EF similarto LD, and similarly 

placed ; therefore EF is given in species, and is about the 

* 26. 6, same diameter'' with LD. Let CBG be 

the diameter, and construct the figure : 

Therefore, because the figure EF given 

in species is described upon the given 

straight lineEB, EFis given in magni- 

' 56. dat. tude% and the gnomon ELH is equal 

* 36. and to the given figure"* ACj wherefore, 

■ ■ since EF is increased by the given gnomon ELH, its sides 
« 82. dat, EK, FH are given-^ j but EK is equal to CD, and FH to BD, 
therefore CD, DB are each of them given. 

This demonstration is the analysis of the problem in the 
aqth Prop. Book 6. the construction and demonstration of 
which is the composition of the analysis. 

Cor. if a parallelogram given in species be applied to a 
given straight line, exceeding by a parallelogram equal to a 
given space ; the sides of the parallelogram are given. 

Let the parallelogram ADCE given in species be applied t© 
the given straight line AB, exceeding by the parallelogram 
BDCG equal to a given space j the sides A D, DC of the pa- 
rallelogram are given. 

Draw 





DATA. 435 

Draw the diameter DE of the parallelogram AGj and con- 
struct the figure. Because the parallelogram AK is equal* to * -^s. 5. 
BC which IS given, therefore AK is 

given ; and BK is similar^ to AC, there- *^^ ^ 5^ '■ -*• 6. 

fore BK is given in species. And since 
the parallelogram AK given in magni- rr 
tude is applied to the given straight line 
AB, exceeding by the parallelogram BK — 
given in species, therefore by this pro- *^ 
position, BD, DK the sides of the excess are given, and the 
straight line AB is given ; therefore the whole AD, as also 
DC, to which it has a given ratio is given. 
PROB. 

To apply a parallelogram similar to a given one to a given 
straight line AB, exceeding by a parallelogram equal to a 
given space. 

To the given straight line AB apply'^ the parallelogram AK ' 29. 6. 
equal to the given space, exceeding by the parallelogram i^K 
similar to the one given. Draw DF, the diameter of BK, 
and through the point A draw AE parallel to BF, meeting 
DF produced in E, and complete the parallelogram AC. 

The parallelogram BC is equal* to AK, that is, to the 
given space; and the parallelogram AC is similar'' to BK; 
therefore the parallelogram AC is applied to the straight line 
AB similar to the one given and exceeding by the parallelo- 
gram BC which is equal to the given space. 

PROP. LXXXV. fr*. 

X F two straight lines contain a parallelogram given 
in magnitude, in a given angle; if the difference of 
the straight hnes be given, they shall each of them 
be given. 

Let AB, BC contain the parallelogram AC given in magni- 
tude, in the given angle ABC, and let the excess of BC above 
AB be given ; each of the straight lines AB, BC is given. 

Let DC be the given excess of BC above 

BA, therefore the remainder BD is equal -pt X*- 

to BA. Complete the parallelogram AD j / / / 

and because AB is equal to BD, the ratio / / / 

of AB to BD is given ; and the angle ABD J- - ^ ^ 

is given, therefore the parallelogram AD is ^ 1> C 

given in species ; and because the given parallelogram AC is 
applied to the given straight line DC, exceedmg by the paral- 
lelogram AD given inspecies, thesides of the excess are given*: • ^- •^'* 

F f 2 therefore 



436 EUCLID'S 

therefore BD is given ; and DC is given, wherefore the whole 
BC is given : and AB is given, therefore AB, BC are each of 
them given. 

«5 PROP. LXXXVI. 

XF two straight lines contain a parallelogram given 
in magnitude, in a given angle ; if both of them to- 
gether be given, they shall each of them be given. 

Let the two straight lines AB, BC contain the parallelogram 
AC given in magnitude, in the given angle ABC, and let AB, 
BC, together be given ; each of the straight lines AB, BC is given . 

Produce CB, and make DB equal to BA, and complete the 
parallelogram ABDE. Because DB is equal to BA, and the 
angle ABD given, because the adjacent an- 
gle ABC is given, the parallelogram AD is - J£___A__ 
given in species : And because AB, BC, to- j 7 7 

gether are given, and AB is equal to BD ; / I 

therefore DC is given: And because the gi- / / / 

yen parallelogram AC is applied to the given 0^ J3 C 
straight line DC, deficient by the parallelo- 
gram AD given in species, the sides AB, BD of the defe£l are 
» 63. dat. given'jand DC is given, wherefore the remainder BC is given j 
and each of the straight lines AB, BC is therefore given. 

87. PROP. LXXXVII. 

IF two straight lines contain a parallelogram given 
in magnitude, in a given angle; if the excess of the 
square of the greater above the square of the lesser 
be given, each of the straight lines shall be given. 

Let the two straight lines AB, BC contain the given paral- 
lelogram AC in the given angle ABC ; if the excess of the 
square of BC above the square of B A be given \ AB and BC 
are each of them given. 

Let the given excess of the square of BC above the square 
of BA be the redangle CB, BD ; take this from the square 
• 2. 2. of BC, the remainder, which is' the re£langle BC, CD is 
equal to the square of AB : and because the a«gle ABC of 
the parallelogram AC is given, the ratio of the redangle 
>«2.iit. of the sides AB, BC to the parallelogram AC is given*" ; and 
AC is given, therefore the reftangle AB, BC is given j and 
the redhngle CB, BD is given \ therefore the ratio of the re£l- 

anglc 



DATA. 



437 




«7.dit. 



angle CB, BD to the rectangle AB, BC, that is% the ratio of \- ^^ 
the straight line DB to BA is given ; therefore"* the ratio ©f *^'*"'^**' 
the square of DB to the square of BA is 
given : And the square of BA is equal to 
the reftangle BC, CD : wherefore the ra- 
tio of the redlangle BC, CD to the square 
of BD is given, as also the ratio of four 
times the rectangle BC, CD to the square 
of ED ; and, by composition*, the ratio of four times the rect- 
angle BC, CD together with the square of BD to the square 
of BD is given : But four times the redtangle BC, CD, toge- 
ther with the square of BD, is equal'' to the square of the straight ' 5- 2- 
lines BC, CD taken together: therefore the ratio of the square 
of BC, CD, together to the square of BD, is given ; wherefore 
sche ratio of the straight line BC, together with CD to BD, is 
given : And, by composition, the ratio of BC together with 
CD and DB, that is, the ratio of twice BC to BD, is given j 
therefore the ratio of BC to BD is given, as also"^ the ratio of 
the square of BC to the rectangle CB, BD : But the re<5langle 
CB, BD is given, being the given excess of the squa-es of BC, 
BA; therefore the square of BC, and the straight line BC, is 
given : And the ratio of BC to BD, as also of BD to BA, has 
been shewn to be given ; therefore*" the ratio of BC to BA is '' '^'* 
given ; and BC is given, wherefore BA is given. 

The preceding demonstration is the analysis of this problem, 



58. dat. 



VIZ. 



A parallelogram AC, which has a given angle ABC, being 
given in magnitude, and the excess of the square of BC, one of 
its sides above the square of the other BA being given ; to 
find the sides : And the composition is as follows : 

Let EFG be the given angle to which the angle ABC is 
required to be equal, and from any point E in F£, draw EG 
perpendicular to FG ; let the reft- 
angle EG, GH be the given space 
to which the parallelogram AC is 
to be made equal ; and the rectangle 
HG, GL, be the given excess of 
the squares of BC, BA. T'~f^ j " A tj~\f 

Take, in the straight line GE, -t ^ Lr ^ -H.-^ 
GK equal to FE, and make GM double of GK ; join ML, 
and in GL produced, take LN equal to LM : Biseft GN in O, 
»nd between GH, GO find a mean proportional BC : As OG 
to GL, so make CB to BD ; and make the angle CBA equal 

F f 3 to 




438 EUCLID'S 

to GFE, and as LG to GK, so make DB to BA, and com- 
plete the parallelogram AC : AC is equal to the reftangle 
EG, GH, and the excess of the squares of CB, BA is equal 
to the reftangle HG, GL. 

Because as CB to BD, so is OG to GL, the square of CB 

* ^' ^' is to the re^angle CB, BD as" the reaangle HG, GO to 
the redlangle HG, GL : and the square of CB is equal to the 
redlangle HG, GO, because GO, BC, GH, are proportionals ; 

' ^*- 5- therefore the reaangle CB, BD is equal" to HG, GL. And 
because as CB to BD, so is OG to GL ; twice CB is to BD, 
as twice OG, that is, GN to GL; and, by division, as BC 
together with Cp is to BD, so is NL, that is, LM, to LG : 

< 22. 6. Therefore^ the square of BC together with CD is to the square 
of BD, as the Square of ML to the square of LG : But the 

*8. 2. square of BC^nd CD together is equal" to four times the 
re£langle BC, CD together with the square of BD ; therefore 
four times the r^angle BC, CD together with the square of 
BD is to the squaYe of BD, as the square of ML to the square 
of LG : And, by division, four times the redtangle BC, CD is 
to the square of BD, as the square of MG to the square of 
GL ; wherefore the rectangle BC, CD is to the square of BD 
as (the square of KG the half of MG to the square of GL, 
that is, as) the square of AB to the square of BD, because as 
LG to GK, so DB was made to BA : Therefore'' the reftan- 
gle BC, CD is equal to the square of AB. To each of these 
add the rectangle CB, BD, and the square of BC becomes equal 
to the square of AB, together with the re6langle CB, BD 
therefore this redtangie, that is, the given rectangle HG, GL,j 
is the excess of the squares of BC, AB. From the point Al 
draw AP perpendicular to BC, and because the angle ABFJ 
is equal to the angle EFG, the triangle ABP is equiangulaij 
to EFG : And DB was made to BA, as LG to GK ; therefom 
as the reaangle CB, BD to CB, BA, so is the redangle HGI 





B PD C F G L HN 



GL to HG, GK ; and as the redangle CB, BA to AP, B( 
so is (the straight line BA to AP, and so is FE or GK 

E( 



DATA. 439 

EG^ and so is) the rectangle HG, GK to HG, GE ; therefore, 
ex jequali, as the rectane;le CB, BD to AP, BC, so is the re<ft- 
angle HG, GL to EG,''GH : And the rectangle CB, BD is 
equal to HG, GL ; therefore the rectangle AP, BC, that is, 
the parallelogram AC, is equal to the given rectangle EG, GH. 



PROP. LXXXVIII. N. 

J F t\ro straight lines contain a parallelogram given 
in magnitude, in a given angle; if the sum of the 
squares of its sides be given, the sides shall each of 
them be given. 

Let the two straight lines AB, BC contain the parallelogram 
ABCD given in magnitude in the given angle ABC, and let 
the sum ofthe squares of AB,BC be given ; AB,BC are each 
of them given. 

First, let ABC be a right angle: and because twice the rect- 
angle contained by two equal straight lines is equal to both 
their squares ; but if two straight lines are un- 
equal, twice the rectangle contained by them 
less than the sum of their squares, as is evident 
from the yth Prop. B. 2. Elem. ; therefore twice 
the given space, to which space the rectangle of which the sides 
are to be found is equal, must not be greater than the given 
sum of the squares of the sides ; And if twice that space be 
equal to the given sum ofthe squares, the sides of the rectangle 
must necessarily be equal to one another : Therefore in this 
case describe a square ABCD equal to the given rectanple, and 
its sides AB, BC are those which were to be found j For the 
rectangle AC is equal to the given space, and the sum ofthe 
squares of its sides AB, BC is equal to twice the rectangle AC, 
that is, by the hypothesis, to the given space to which the suna 
of the squares was required to be equal. 

But if twice the given rectangle be not equal to the given 
sum of the squares of the sides, it must be less than it, as 
has been shown. Let ABCD be the rectangle, join i\C arid 
draw BE perpendicular to it, and complete the rectangle 
AEBF, and describe the circle ABC about the triangle ABC ; 
AC is its diameter-i : And because the triangle ABC is simi- * C»r.5. ♦. 
lar^ to AEB, as AC to CB, so is AB to BE ; therefore the " 8. 6. 
recungle AC, BE is equal to AB, BC i and the rectangle AB, 

Ff4 ' BC 



ce B' 'C 




^o E U C L I I>' S 

BC is given, wherefore AC, BE is given : And because the sum 

of the squares of AB, BC is given, the square of AC which is 

-47, 1, equal"^ to that sum is given j and AC itself is therefore given 

ir> magnitude ; Let AC be liicewise given in position, and the 

* 32. dat. point A J therefore AF is given'' in po- 

sition : And the rectangle AC, BE is 

given, as has been shewn, and AC is 
' 61 .dat. given, wherefore^ BE is given in mag-» 

nitude, as also AF which is equal to it ; 

and AF is also given in position, and 
fso. dat. the point A is given, wherefore*^ the 

point F is given, and the straight line ^ — -r^ ls~f 

* 3i.dat. FB in positions; And the circumfe- ^' ^ H 1^ 
•"SS. dat, rence ABC is given in position, wherefore'' the point B is 

given 1 And the points A, C are given ; therefore the straight 

5 29.dat. lines AB, BC are given' in position and magnitude. 

The sides AB, BC of the rectangle maybe found thusj Let 
the rectangle GH, GK be the given space to which the rect- 
angle AB, BC is equal; and let GH, GL be the given rect- 
angle to which the sum of the squares of AB, BC is equal : 

" 14. 2. Find ^ a square equal to the rectangle GH, GL ; And let its 
side AC be given in position; upon AC as a diameter describe 
the semicircle ABC, and as AC to GH, so make GK to AF, 
and from the point A place AF at right angles to AC : There- 

116. 6. fore the rectangle CA, AF is equal ' to GH, GK ; and, by 
the hypothesis, twice the rectangle GH, GK is less than GH, 
GL, that is, than the square of AC; wherefore twice the 
rectangle CA, AF is less than the square of AC, and the rect- 
angle CA, AF itself less than half the square of AC, that is, 
than the rectangle contained by the diameter AC and its half; 
wherefore AF is less than the semidiameter of the circle, and 
consequently the straight line drawn through the point F pa- 
rallel to AC must meet the circumference in two points: Let 
B be either of them, and join AB, BC, and complete the rect- 
angle ABCD, ABCD is the rectangle which was to be found: 

p" 34. 1. Draw BE perpendicular to AC ; therefore BE is equal »" to 
AF, and because the angle ABC in a semicircle is a right an-- 

* 8. 6. gle, the rectangle AB, BC is equal'' to AC, BE, that is, to 

the rectangle CA, AF which is equal to the given rectangle 

^47. 1. GH, GK: And the squares of AB, BC are together equal*" tO( 

the square of AC, that is, to the given rectangle GH, GL 

But if the given angle ABC of the parallelogram AC be not 
a right angle, in this case, because ABC is a given angle, th«f- 
ratio of the rectangle contained by the sides AB, BC to the pa- 

rallelograin 



DATA. 



44» 



rallelogram AC is given* j and AC is given, therefore the * 62. dau 
rectangle AB, BC is given : and the sum of the squares of 
AB, BC is given ; therefore the sides AB, BC are given by 
the preceding case. 

The sides AB, BC, and the parallelogram AG, may be found 
thus : Let EFG be the given angle of the parallelogram, and 
from any point E in FE draw EG perpendicular to FG : and 
let the rectangle EG, FH be the given space to which the pa- 
rallelogram is to be made equal, and let EF . .^ 

FK be the given rectangle to which the -p= 4-' 

s una of the squares of the sides is to be equal . / / 

And, by the preceding case, find the sides _ / J. 



of a rectangle which is equal to the given O 1^ 
rectangle EF, FH, and the squares of the 
sides of which are together equal to the ] 

given rectangle EF, FK ; therefore, as was 
shewn in that case, twice the rectangle EF, 
FH must not be greater than the rectangle 
EF, FK: let it be so, and let AB, BC be 
the sides of the rectangle jo'ned in the 
angle ABC equal to the given angle EFG, 



c 



F llGr K 



and complete the parallelogram ABCD, which will be that 
which was to be found : Draw AL perpendicular to BC, and 
because the angle ABL is equal to EFG, the triangle ABL is 
equiangular to EFG ; and the parallelogram AC, that is, the 
rectangle AL, BC, is to the rectangle AB, BC as (the straight 
line AL to AB, that is, as EG to EF, that is, as) the rectan- 
gle EG, FH to EF, FH ; and, by the construction, the rect- 
angle AB, BC is equal to EF, FH, therefore the rectangle AL, 
BC or, its equal, the parallelogram AC, is equal to the given 
rectangle EG, FH ; and the squares of AB, BC are together 
equal, by construction, to the given rectangle EF, FK. 



442 E U C L I D'S 

96. PROP. LXXXIX. 

XF two straight lines contain a given parallelogram 
in a given angle, and if the excess of the square of 
one of them above a given space, has a given ratio 
to the square of the other ; each of the straight 
lines shall be given. 

Let the two straight lines AB, BC contain the given paral- 
lelogram AC in the given angle ABC, and let the excess of the 
square of BC above a given space have a given ratio to the 
square of AB, each of the straight lines AB, BC is given. 

Because the excess of the square of BC above a given space 
has a given ratio to the square of BA, let the rectangle CB, 
BD be the given space; take this from the square of BC, the 

■' 2. 2. remainder, to wit, the rectangle* BC, CD has a given ratio to 
the square of BA : Draw AE perpendicular to BC, and let the 
square of BF be equal to the rectangle BC, CD, then, because 
the angle ABC, as also BEA is given, the 

*4j. dat. triangle ABE is given ''in species, and the ly 



ratioof7\E to ABisG;iven: Andbecause the ■^/— 



/ 



ratioof the rectangle BC, CD, that is, of 

the square of BF to the square of BA, is -ij— f^^ 7^ 

given, the jatio of the straight line BF to ^^ ^ 
"^.08. dat, BA is given"= and the ratio of AE to AB is given, wherefore'- 
" ^* ^"*" the ratio of A£ to BF is given ; as also the ratio of the rcct- 
''^^- ^- angle AE, BC, that is% of the parallelogram AC to the rect- 
angle FB, BC ;and AC is given, wherefore the redtangle FB, 
BU'is given. The excess of the square of BC above the 
square of BF, that is, above the redlangle BC, CD, is given, 
for it is equal=^ to the given redtangleCB. BD ; therefore, be- 
cause the reftangle contained by the straight lines FB, BC is 
given, and also the excess of the square of BC above the square 
'87. <ijt. of BF ; FB, BC,are each of them given*"; and the ratio of FB 
to BA is given ; therefore AB, BC are given. 

The Composition is as follows : 

Let GHK be the given angle to which the angle of the 
paralk-logram is to be made equal, and from any point G in 
HG, draw GK perpendicular tQ HK; let GK, HL be the 

re<5langle 



443 



DATA. 

rectangle to which the parallelogram is to be 
made equal, and let LH, HM be the rect- 
angle equal to the given space which is to 
be taken from the square of one of the sides ; 
and let the ratio of the remainder to the t/ tI '\f- 
square of the other side be the same with the -*^ ^ *• *-• 

ratio of the square of the given straight line NH to the square 
of the given straight line HG. 

By help of the 87th dat. find two straight lines BC, BF, 
which contain a rectangle equal to the given rectangle NH, 
HL, and such that the excess of the square 
of BC above the square of BF be equal to 
the given rectangle LH, HM j and join 
CB, BF in the angle FBC equal to the 

given angle GHK : And as NH to HG, so ^„ ' ^/ 

make FB to BA,and complete the paralle- ^ j^^ l) C- 
logram AC, and draw AE perpendicular to BC : then AC is 
equal to the rectangle GK, HL ; and if from the square of < 
BC, the given rectangle LH, HM be taken, the remainder 
shall have to the square of B A the same ratio which the square 
of N H has to the square of HG. 

Because, by the construction, the square of BC is equal to 
the square of BF together with the rectangle LH, HM ; if 
from the square of BC there be taken the rectangle LH, HM, 
there remains the square of BF, which has s to the square of e 22. c. 
BA the same ratio which the square of NH has to the square 
of HG, because, as NH to HG, so FB was made to BA : but 
as HG to GK, so is BA to AE, because the triangle GHK is 
equiangular to ABE j therefore, ex aequali, as NH to GK, so is 
FB to AE ; wherefore ^ the rectangle NH, HL is to the rect- h 1. g.. 
angle GK, HL, as the rectangle FB, BC to AE, BC ; but by 
the construction the rectangle NH, HL is equal to FB, BC ; 
therefore* the rectangle GK, HL is equal to the rectangle i 14 5 
AE, BC, that is, to the parallelogram AC. 

The analysis of this problem might have been made as in 
the 86th Prop, in the Greek, and the composition of it may be 
made as that which is in Prop. 87th of this edition. 



444 E U e L I D'S 

PROP. XC. 
o. 

XF two straight lines contain a given parallelo- 
gram in a given angle, and if the square of one of 
them together with the space, vvhich has a given 
ratio to the square of the other be given, each of 
the straio'ht lines shall be'ii'iven. 

lyst the two straight lines AB, BC contain the given pa- 
rallelogram AC in the given angle ABC, and let the square 
of BC together with the space which has a given ratio to the 
square of AB be given, AB, BC are each of them given. 

Let the square of BD be the space which has the given ratio 

to the square of AB ; therefore, by the hypothesis, the square 

of BC together with the square of BD is given. From the 

point A, draw AE perpendicular to BC ; and because the angles 

, 43. dst. ABE, BEA are given, the triangle ABE is given* in species ; 

therefore the ratio of BA to AE is given : And because the 

ratio of the square of BD to the square of BA is given, the 

* 58. dat. ratio of the straight line BD to BA is given'' j and the ratio of 

" 9. dat. BA to AE is given j therefore'^ the ratio of AE to BD is 

giv*en, as also the ratio of the re£lang\e AE, BC, that is, of the 

parallelogram AC to the rectangle DB, BC i and AC is given, 

therefore the redangle DB, BC is given; and the square of 

1> 



aZ 



B E C 



GH K L 



J ,j ,,^^ BC together with the square of BD is given :» therefore'^ be- 
cause the re6tangle contained by the two straight lines DB, 
I>C is given, and the sum of their squares is given: The 
!Xtraight lines DB, BC are each of them given; and the ratio 
of DB to BA is given ; therefore AB, BC are given. 

The Compos ilioji is as follows : 

Let FGH be the given angle to which the angle of the 
parallelogram is to be made equal, and from any point F in GF 
draw FH perpendicular to GH ; and let the rectangle FH, 
GK be that to which the parallelogram is to be made equal; 
and let the redangle KG, GL be the space to which the square 

of 



DATA. 

«f one of the sides of the parallelogram, together with the space 
which has' a given ratio to the square of the other side, is to be 
made equal ; and let this given ratio be the same which the 
square of the given straight line MG has to the square of GF. 
By the 88th dat. find two straight lines DB, BC which con- 
tain a redangle equal to the given redangle MG, GK, and 
such that the sum of their squares is equal to the given revS- 
angle KG, GL ; therefore, by the determination of the pro- 
blem in that proposition, twice the redangle MG, GK must 
not be greater than the re^an^le KG, GL. Let it be so, and 
jom the straight lines DB BC^in the an^e D3C equal to the 
given angle FGH ; and, as MG to GF^, so make DB,to BA, 
and complete the parallelogram AC : AC is equal to the rea- 

A/ 



445 



/rj 



HE C ^^ K L 

angle FH, GK ; and the square of BC together with the square 
of BD, which, by the construetion, has to the square of BA 
the given ratio which the square of MG has to the square of 
GF, is equal, by the construction, to the given rectangle 
KG, GL. Draw AE perpendicular to BC. 

Because, as DB to BA, so is MG to GF ; and as BA to 
AE, so GF to FH ; ex aequali, as DB to AE, so is MG to 
FH i therefore, as the rectangle DB, BC to AE, BC, so is 
the rectangle MG, GK to FH, GK ; and the rectangle DB, 
BC is equal to the rectangle MG, GK ; therefore the rectan- 
gle AE, BC, that is the parallelogram AC, is equal to the 
rectangle FH, GK. 

PROP. XCL 

J.F a straight line drawn within a circle givTn in 
magnitude cuts off a segment which contains a gi- 
ven angle; the straight line is given in magnitude. 

In the circle ABC given in magnitude, let the straight line 
AC be drawn, cutting off the segment AEC which contains the 
given angle AEC ; the straight line AC is given in magnitude. 

Take D the centre of the circle*, join AD, and produce it ■. , 

to 




446 K U C L I D*5 

to E and join EC : The angle ACE being 

* 31. 3. a right'' angle, is given ; and the angle 
' *3- dat. ^EC is given i therefore'^ the triangle 

ACE is given in species, and the ratio of 
EA to AC is therefore given, and EA is 
given in magnitude, because the circle is ^ 
d 5 jgf^ given"* in magnitude j AC is therefore 

* 2. dat. given'' in magnitude. 

PROP. XCII. 

V a straight line given in magnitude be drawn 
within a circle given in magnitude, it shall cut off 
a segment containing a given angle. 

Let the straight line AC given in magnitude be draw^n 
within the circle ABC given in magnitude j it shall cut off 
a segment containmg a given angle. 

Take D the centre of the circle, join 
AD and produce it to E, and join EC : 
And because each of the straight lines EA 
and AC is given, their ratio is given^ : and 
the angle ACE is a right angle, therefore ^ 
the triangle ACE is given** in species, 
and consequently the angle AEC is given. 

PROP. XCIII. 

j[ F from any point in the circumference of a circle 
given in position two straight lines be drawn, meet- 
ing the circumference and containing a given angle ; 
if the point in which one of them meets the cir- 
cumference again be given, the point in which the 
other meets it is also given. 

From any point A in the circumference of a circle ABC 
given in position, let AB, AC be drawn to the circumference 
making the given angle BAC -, if the point ^,— -^ 

B be given, the point C is also given. 

Take D the centre of the circle, and 
join BD, DC j and because each of the 
points B, D is given, BD is given* in g\ 
position i and because the angle BAC is 
given, the angle BDC is given'', therefor? 

I because 



» 1. dat. 
> 46. dat. 

90. 




» 29, dat. 
» 'JO. 3. 




DATA. 447 

because the straight line DC is drawn to the given point D in 
the straight line BDgiven inposition in the given angle BDC, 
DC is given'^ in position: And the circumference ABC is '■^-°^-- 
given in position, therefore"^ the point C is given. * '^^ 

PROP. XCIV. .«!. 

J-F from a given point, a straight line be drawn 
touching a circle given in position ; the straight 
line is given in position and magnitude. 

Let the straight line AB be drawn from the given point 
A, touching the circle BC given in position ; AB is given in 
position and magnitude. 

Take D the centre of the circle, and join DA, DB : Because 
each of the points D, A is given, the 
straight line AD is given* in position 
and magnitude: And DBA is aright*" 
angle, wherefore DA is a diameter"^ of f 
the circle DBA, described about the 
triangle DBA ; and that circle is there- 
fore given*^ in position : And the circle \^^ ^ ^ *^-^^'- 
BC is given in position, therefore the 

point B is given^ The point A is also given: Therefore e-^^ 
the straight line AB is given* in position and magnitude. 

PROP. XCV^. 9^2. 

IF a straight Hue be drawn from a given point 
without a circle given in position ; the rectangle 
contained by the segments betwixt the point and 
'the circumference of the circle is eiveii. 

o 

Let the straight line ABC be drawn from the given point A 
without the circle BCD given in po- 
sition, cutting it in B, C j the rectan- 
gle BA, AC is given. 

From the point A, draw* AD ^.. 
touching the circle J therefore AD is V /ij-A. 

given'' in position and magnitude : \ y " ^- ^^^ 

And because AD is given, the square ^ ^ 

of AD is givenS which is equaW to the rectangle BA *'^ ' 56. dat. 
Therefore the rectangle BA, AC is given 






448 E U C L I D'S 

93. ■ PROP. XCVI. 

If a straight line be drawn through a given point 
within a circle given in position, the rectangle con- 
tained by the segments betwixt the point and the 
circumference of the circle is given. 

Let the straight line BAG be drawn through the given point 
A within the circle BCE given in position j the rectangle 
BA, AC is given. 

Take D the centre of the circle, join 
AD, and produce it to the points E, F : 
Because the points A, D are given, the 

» 29. dat. straight line AD is given* in position ; 
and the circle BEC is given in position ; 

» 28. dat. therefore the points E, F are given''} and B^ 
the point A is given, therefore EA, AF 
are each of them givan% and the re£t- 

*35. 3. angle Ey\, AF is therefore given j and it is equal'^ to the re6l« 
angle BA, AC, which consequently is given. 

9*- PROP. XCVII. 

xF a straight line be drawn within a circle given 
in magnitude, cutting off a segment containing a 
given angle ; if the angle in the segment be bisected 
by a straight line produced till it meets the cir- 
cumference, the straight lines which contain the 
given angle shall both of them together have a gi- 
ven ratio to the straight line which bisects the an- 
gle. And the rectangle contained by both these 
lines together which contain the given angle, and 
the part of the" bisecting line cut off below the base 
of the segment, shall be given. 

Let the straight line BC be drawn within the circle ABC 
given in magnitude, cutting off a segment containing the gi- 
ven angle BAC, and let the angle BAC be bisected by the 
straio-ht line AD ; BA together with AC has a given ratio to 
AD ; and the rectangle contained by BA and AC together, 
and the straight line ED cut off from AB below BC the base 
of the segment is given. 

Join BD j and because BC is drawn within the circle ABC 
2 given 




DATA. 449 

»iven in magnitude cutting off the segment BAC, containing 
the given angle BAC ; BC is given'" in magnitude : By the * ^i- <»»*• 
same reason BD is given ; therefore'' the ratio of BC to BD " i- dat, 
is given : And because the angle BAC is bise^ed by AD, as 
BA to AC, so is'' BE to EC; and, by permutation, as AB*3.6. 
to BE, so is AC to CE j wherefore^ as BA and AC together ' 12. 5. 
to BC, so is AC to CE : And because the angle B AE is equal 

to EAC, and the angle ACE to' -p,^ ^ ^ ^ '21. 

ADB, the triangle ACE is equian- 
gular to the triangle ADB; therefore 
as AC to CE, so is AD to DB : But 
as AC toCE, so is BA together with 
AC to BC; as therefore BA and AC 
to BC, so is AD to DB : and, by 
permutation, as BA and AC to AD 

so is BC to BD ; And the ratio of BC to BD is given, there- 
for* the ratio of B A together with AC to AD is given. 

Also the redtangle contained by BA and AC together, and 
DE is given. 

Because the triangle BDE is equiangular to the triangle 
ACE, as BD to DE, so is AC to CE ; and as AC to CE, so 
is BA and AC to BC; therefore as BA and AC to BC, so is 
BD to DE ; wherefore the reitangie contained by BA and 
AC together, and DE, is equal to the re(Sangle CB, BD : 
But CB, BD is given ; therefore the rei^angle contained by 
BA and AC together, and DE, is given. 



Otherwhe^ 

Produce CA, and make AF equal to AB, and join BF ; 
and because the angle BAC is double^ of each of the angles *>32. j. 
BFA, BAD, the angle BFA is equal to BAD ; and the angle 
BCA is equal to BDA, therefore the triangle FCB is equian- 
gular to ABD : As therefore FC to CB, so is AD to Y>^ ; 
and, by permutation, as FC, that is, BA and AC together, to 
AD, so is CB to BD : And the ratio of CB to BD is given, 
therefore the ratio of BA and AC to AD is given. 

And because the angle BFC is equal to the angle DAC, 

that is, to the angle DBC, and the angle ACB equal to th4 

^angle ADB; the triangle FCB is equiangular to BDE, as 

"therefore FC to CB, so is BD to DE ; therefore the Fedbngle 

contained by FC, that is, BA and AC together, and DE is 

G g equal 



450 EUCLID'S 

equal to the re<ftangle CB, BD which is given, and therefore 
the redangie contained by BA, AC together, and-DE is given. 



F. PROP. XCVIII. 

IF a straight line be drawn within a circle given in 
magnitude, cutting ofFa segment containing a given 
angle : If the angle adjacent to the angle in the seg- 
ment be bisected by a straight line produced till it 
meet the circumference again, and the base of the 
segment ; the excess of the straight lines which con^ 
tain the given angle shall have a given ratio to the 
segment of the bisecting line which is within the 
circle ; and the rectangle contained by the same ex- 
cess, and the segment of the bisecting line betwixt 
the base produced and the point where it again meets 
the circumference, shall be given. 

Let the straight line BC be drawn within the circle ABC 
given in magnitude, cutting ofFa sejTment containing the given 
angle BAC, and let the angle CAF adjacent to BAC be bi- 
seiSed by the straight line DAE, meeting the circumference 
again in D, and BC the base of the segment produced in E ; 
the excesss of BA, AC has a given ratio to AB ; and the 
re<5langle which is contained by the same excess and the straight 
line EJD is given. 

Join BD, and through B, draw BG parallel to DE meeting 
AC produced in G: And because BC cuts off from the circle 
ABC given in magnitude the seg- 
ment BAC containing a given an- 
•*i. dat., gle, BC is therefore given" in mag- 
nitude : By the same reason BD is 
given, because the angle BAD is 
equal to the given angle EAF ; 
therefore the ratio of BC to B D is 
given : And because the angle CAE 
IS equal to EAF, of vyhich CAE ^^r, 

is equal to the alternate angle AGB, and EAF to the interior 
and opposite angle ABG j therefore the angle AGB is equal to 
ABG, and the straight line AB equal to AG j so that GC is 

the 




DATA. 451 

the excess of BA, AC: And because the angle BGC is equal 
to GAE, that is to EAF, or the angle BAD ; and that the 
angle BCG is equal to the opposite interior angle BDA of 
the quadrilateral BCAD in the circle ; therefore the triangle 
BGC is eqHiangular to BDA . Therefore as GC to CB, so 
is AD to DB : and, by permutation, as GC which is the ex- 
cess of BA, AC to AD, so is BC to BD : And the ratio of 
CB to BD is given ; therefore the ratio of the excess of BA, 
AC to AD is given. 

And because the angle GBC is equal to the alternate angle 
DEB, and the angle BCG equal to BDE ; the triangle BCJG 
is equiangular to BDE ; Therefore as GC to CB, so is BD 
to DE : and consequently the rectangle GC, DE is equal to 
the rectangle CB, BD which is given, because its sides CB, 
BD are given ; Therefore the rectangle contained by the ex- 
cess of BA, AC and the straight line DE is given. 



PROP. XCIX. 

J-F from the given point in the diameter of a circle 
given in position, or in the diameter produced, a 
straight line be drawn to any point in the circumfe- 
rence, and from that point a straight line be drawn 
at right angles to the first, and from the point in 
which this meets the circumference again, a straight 
line be drawn parallel to the first ; the point in 
which this parallel meets the diameter is given; and 
the rectangle contained by the two parallels is given. 

In BC the diameter of the circle ABC given in position, or 
in BC produced, let the given point D be taken, and from D 
let a straight line DA be drawn to any point A in the circum- 
ference, and let AE be drawn at right angles to DA, and from 
the point E where it meets the circumference again let EF 
be drawn parallel to DA meeting BC in F ; the point F is gi- 
ven, as also the rectangle AD, EF. 

Produce EF to the circumference in G, and join AG : 
because GEA is a right angle, the straight line AG is* the 
diameter pf the circle ABC ; and BC is also a diameter of it; 
therefore the point H, where they meet, is the centre of the 
circle, and consequently H is given : And the point Dis given, 
wherefore DH is given in magnitude. And because AD is 

G g 2 parallel 



95. 



» Cor. 5. 4. 



4S2 

»> 4. 6. 



E U C L I D'S 



parallel to FG, and GH equal to HA ; DH is equal'' to HF, 
and AD equal to GF : And DH is given, therefore HF is givea 



« 30. dat. 



(J 95. or 96, 
dat. 





J>J 



in magnitude ; and it is also given in position, and the poinfc 
H is given, therefore<= the point F is given. 

And because the straight line EFG is drawn from a given 
point F without or within the circle ABC given in position^ 
therefore^ the rectangle EF, FG is given : And GF is equal 
to AD, wherefore the rectangle AD, EF is given. 

PROP. C. 

If from a given point in a straight line given in 
position, a straight line be drawn to any point in 
the circumference of a circle given in position ; and 
from this point a straight line be drawn, making 
with the first an angle equal to the difference of a 
right angle, and the angle contained by the straight 
line given in position, and the straight line ^rhich 
joins the given point and the centre of the circle ; 
and from the point in which the second line meets, 
the circumference again, a third straight line be 
drawn, making with the second an angle equal to 
that which the first makes with the second : The 
point in which this third line meets the straight 
line given in position is given; as also the rectan- 
gle contained by the first straight line and the seg- 
ment of the third betwixt the circumference and 
the straight hne given in position, is given. 

Let the straight line CD be drawn from the given point C, 
in the straight line AB given in position, to the circumference 
of the circle DEF given in position, of which G is the centre; 
join CG, and from the point D let DF be drawn, making the 
angle CDF equal to the diflFerence of a right angle, and the 
angle BCG, and fronti the point F let FE be drawn, making 

the ., 



453 



'26.3. 



DATA. 

the angle DFE equal to CDF, meeting AB in H : The point H 

is given j as also the rectangle CD, 

FH. 

Let CD, FH n>eet one another in 
the point K, from which draw KL 
perpendicular to DF j and let DC 
meet the circumference again in M, a\ 
and let FH meet the same in E, and 
join MG, GF, GH. 

Because the angles MDF, DFE 
are equal to one another, the circum- 
ferences MF, DE are equal* ; and 
adding or taking away the common 
part ME, the circumference DM is 
equal to EF; therefore the straight 
line DM is equal to the straight line 
EF, and the angle GMD to the an- 
gle'' GFE ; and the angles GxMC, 
GFH are equal to one another, be- 
cause they are either the same with 
the angles GMD, GFE, or adjacent 
to them : And because the angles 
KDL, LKD are together equal"^ to 
a right angle, that is, by the hypo- 
thesis, to the angles KDL, GCB ; 
the angle GCB or GCH is equal to 
the angle (LKD, that is, to the angle) LKF or GKH : There- 
fore the points C, K, H, G are in the circumference of a cir- 
cle J and the angle GCK is therefore equal to the angle GHF; 
and the angle GMC is equal to GFH, and the straight line 
GM to GF ; therefore'' CG is equal to GH, and CM to HF : *26. i. 
And because CG is equal to GH, the angle GCH is equal to 
GHC i but the angle GCH is given : Therefore GHC is gi- 
ven, and consequently the angle CGH is given ; and CG is 
given in position, and the point G ; therefore' GH is given in 
position ; and CB is also given in position, wherefore the *" *** 
point H is given. 

And because HF is equal to CM, the rectangle DC, FH is 
equal to DC, CM : But DC, CM is given^ because the point-^^^ »' ^c. 
C is given, therefore the rectangle DC, FH is given. 




"S.l. 



«32. 1. 



dat. 



END OF THE DATA. 

Gg3 



455 



NOTES 

ON 

EU C LID'S DATA 



DEFINITION II. 

X HIS is made more explicit than in the Greek text, to pre- 
vent a mistake which the author of the second demonstra- 
tion of the 24th Proposition in the Greek edition has fallen 
into, of thinking that a ratio is given to which another ratio is 
shewn to be equal, though this other be not exhibited in given 
magnitudes. See the Notes on that Proposition, which is the 
13th in this edition. Besides, by this definition, as it is now 
given, some propositions are demonstrated, which in the Greek 
are not so well done, by help of Prop. 2. 

DEF. IV. 

In the Greek text, def. 4. is thus : " Points, lines, space?, 
** and angles are said to be given in position which have always 
** the same situation ;" but this is imperfectand useless, because 
there are innumerable cases in which things maybe given ac- 
cording to this definition, and yettheir position cannot be found; 
for instance, let the triangle ABC be given in position, and let 
it be proposed to draw a straight line BD from the angle at B 
to the opposite side AC, which shall cut 
off the angle DBC, which shall be the -A. 

seventh part of the angle ABC; suppose ^y^^ \^ 

this is done, therefore the straight line ^x'''''____— — — ^V^ 
BD is invariable in its position, that is t » "" ^ — ^ 

has always the same situation ; for any 

other straight line drawn from the point B on either side of 
BD cuts off an angle greater or lesser than the seventh part of 
the angle ABC; therefore, according to this definition, the 
straight line BD is given in position, as also* the point D in »28. dat 
which it meets the straight line AC which is given in position. 
But from the things here given, neither the straight line BD 
nor the point D caa be found by the help of Euclid's Ele- 
ments ool^, by which every thing in his data is supposed may 

Gg4 be 



456 N O T E S O N 

be found. This definition is therefore of no use. We have 
amended it by adding, " and which are either actually exhi- 
" bited- or can be found ;" for nothing is to be reckoned gi- 
ven, which cannot be found, or is not actually exhibited. 

The definition of an angle given by position is taken out of 
the 4th, and given- more distinctly by itself in the definition 
marked A. 

DEF. XI. Xli: XIII. XIV. XV. 

The nth and 12th are omitted, because they cannot be 
given in English so as to have any tolerable sense ; and there- 
fore, wherever the terms defined occur, the words which ex- 
press their meaning are made use of in their place. 

The 13th, i4thy 15th are omitted, as being of no use. 

It is to be pbserved in general of the data in this book, that 
t)iey are to be understood to be given geometrically, not al- 
ways arithmetically, that is, they cannot always be exhibited 
in numbers ; for instance, if the side- of a square be given, 
*44. dat. the ratio of it.to its diameter is given'' geometrically, but not 
3. dat. i^ numbers ; and the diameter is given*^; but though the num- 
ber of any equal parts in the side be given, for example 10, 
the number of them in the diahieter cannot be given : And the 
like holds in many other cases^ .j j^ \y., rO ortJ I'X 

PROPOSITION I. 

In this it is shewn that A is to B, as C to D, from this, that 
A is to C, as B to D, and then by permutation ; but it fol- 
lows directly without these two steps, from 7. to 5. 

>,n v.u> . PROP. II. 

The limitation added at the end of this proposition between 
the inverted commas is quite necessary, because without it the 
proposition cannot always be demonstrated : For the author 
having said*, " because A is given, a magnitude equal to it 
« 1. def. " can be found* : let this be C ; and because the ratio of A to 
*>Q. def. " B is given, a ratio which is the same to it can be found V 
addsj " let it be found, and let it be the ratio of C to a." 
Now, from the second definition, nothing more follows 
than that some ratio, suppose the ratio of E to Z, can be 
found, which is the same with the ratio of A to B ; and 
when the author supposes that the ratio of C to A, which is 

also 

* See Dr. Gregory's ^editba of the Dato, 



i 



E U C L I D'S D A T A. 4.57 

also the same with the ratio of A to B, can be found, he ne- 
cessarily supposes that to th,e three magnitudes E, Z, C, a 
fourth proportional A may be found; but this cannot always 
be done by the Elements of Euclid ; from which it is plaia 
Euclid must have understood the proposition under the limita- 
tion which is now added to his vext. An example will make 
this clear : Let A be a given angle, 
and B another angle to which A has 
a given ratio ; for instance, the ratio 
of the given straight line E to the 
given one Z ; then, having found an 
angle C equal to A, how can the an- 
gle A be found to which C has the C -p . 
same ratio that E has to Z ? Cer- 




tainly no way, until it be shown / \ Z 
how to find an angle to which a / \ 

given angle has a given ratio, which 

cannot be done by Euclid's Elements, nor probably by any 
Geometry known in his time. Therefore, in all the proposi- 
tions of this book which depend upon this second, the above- 
mentioned limitation must be understood, though it be not 
explicitly mentioned. 

PROP. V. 

The order of the Propositions in the Greek text between 
Prop. 4. and Prop. 25. is now changed into another which is 
more natural, by placing those which are more simple before 
those which are more complex ; and by placing together those 
which are of the same kind, some of which were mixed among 
others of a different kind. Thus, Prop. 12. in the Greek is 
now made the 5th, and those which were the 22d and 23d 
are made the 1 ith and 12th, as they are more simple than the 
propositions concerning magnitudes, the excess of one of 
which above a given magnitude has a given ratio to the other, 
after which these two were placed ; and the 24th in the Greek 
text is, for the same reason, made the 13th. 

PROP. VI. VII. 

These are universally true, though, in the Greek text, 
they are demonstrated by Prop. 2. which has a limitation ; 
.they are therefore now shewn without it. 



458 N O T E S O N 



PROP. XIL 



In the 23d Prop, in the Greek text, which here is the 12th, 
the words, " /x*! rer auras b»," are wrong translated by Claud. 
Hardy, in his edition of Euclid's Data, printed at Paris, anno 
1625, which was the first edition of the Greek text; and Dr. 
Gregory follows him in translating them by the words, " etsi, 
*' non easdem," as if the Greek had been n v.xt (/.-n Im »vrus 
as in Prop. 9. of the Greek text. Euclid's meanmg is, that 
the ratios mentioned in the proposition must not be the same ; 
for, if they were, the proposition would not be true. What- 
ever ratio the whole has to the whole, if the ratios of the parts 
of the first to the parts of the other be the same with this ra- 
tio, one part of the first may be double, triple, &c. of the 
other part of it, or have any other ratio to it, and consequently 
cannot have a given ratio to it j wherefore, these words must 
be rendered by " non autem, easdem," but not the same ratios, 
as Zambertus has translated them in his edition. 



PROP. XIII. 

Some very ignorant editor has given a second demonstration 
of this proposition in the Greek text, which has been as igno- 
rantly kept in by Claud. Hardy and Dr. Gregory, and has been 
retained in the translations of Zambertus and others ; Carolus 
Renaldinus gives it only : The author of it has thought that 
a ratio was given, if another ratio could be shewn to be the 
same to it, though this last ratio be not found : But this is al- 
together absurd, because from it would be deduced that the ratio 
of the sides of any two squares is given, and the ratio of the di- 
ameters of any two circles, &c. And it is to be observed, that 
the moderns frequently take given ratios, and ratios that are 
always, the same, for one and the same thing } and Sir Isaac 
Newton has fallen into this mistake is the 17th Lemma of his 
Principia, edit. 1713, and in other places i but this chould be 
carefully avoided, as it may lead into other errors. 

PROP. XIV. XV. 

Euclid in this book has several propositions concerning 
magnitudes, the excess of one of which above a given magni- 
tude 



E U C L I D'S D A T A. 459 

tude has a given ratio to the other ; but he has given none 
concerning magnitudes whereof one together with a given 
magnitude has a given ratio to the other ; thousih thess last 
occur as frequently in the solution of problems as the first; the 
reason of which is, that the last may be all demonstrated by help 
of the first ; for it a magnitude, together with a given magni- 
tude, has a given ratio to another magnitude, the excess of 
this other above a given magnitude shall have a given ratio to 
the first, and on the contrary ; as we have demonstrated in 
Prop. 14 And for a like reason, Prop. 15. has been added 
to the Data. One example will make the thing clear : Sup- 
pose it were to be demonstrated, that if a magnitude A toge- 
ther with a given magnitude has a given ratio to another mag- 
nitude B, that the two magnitudes A and B, together with a 
given magnitude, have a given ratio to that other magnitude 
B ; which is the same proposition with respect to the last kind 
of magnitudes above-mentioned, that the first part of Prop. 
16, in this edition is in respectof the first kind : This is shewn 
thus, from the hypothesis, and by the first part of Prop. 14. 
the excess of B above a given magnitude has unto A a given 
ratio; and, therefore, by the first part of Prop. 17. the excess 
of B above a given magnitude has unto B and A together a 
given ratio ; and by the second part of Prop. 14. A and B to- 
gether with a given magnitude has unto B a given ratio; 
which is the thing that was to be demonstrated. In like man- 
ner, the other propositions concerning the last kind of mag- 
nitudes may be shewn. 

PROP. XVI. XVII. 

In the third part of Prop. i®. in the Greek text, which Is 
the i6th in this edition, after the ratio of EC to CB has been 
shown to be given; from this, by inversion and conversion 
the ratio of BC to BE is demonstrated to be given; but with- 
out these two steps, the conclusion should have been made only 
by citing the 6th Proposition. And in like manner, in the 
first part of Prop. 11. in the Greek, which in this edition is 
the 17th from the ratio of DB to BC being given, the ratio of 
DC to DB is shewn to be given, by inversion and composi- 
tion, instead of citing Prop. 7. and the same fault occurs in 
the second part of the same Prop. 11. 



46(5 N O T E S O N 

PROP. XXI. XXII. 
These now are added, as being wanting to complete the 
subject treated of in the four preceding propositions. 

PROP. XXIII. 

This which is Prop. 20. in the Greek text, was separated 
from Prop. 14. 15. ib. in that text, after which it should have 
been immediately placed, as being of the same kind ; it is now 
put into its proper place j but Prop. 21. in the Greek is left 
out, as being the same with Prop. 14. in that text, which is 
here Prop. 18. 

PROP. XXIV. 

This, which is Prop. 13. in the Greek, is now put into its 
proper place, having been disjoined from the three follow^ing 
it in this edition, which are of the same kind, 

PROP. XXVIII. 

This, which in the Greek text is Prop. 25. and several of 
the following propositions, are there deduced from Def. 4. which 
, is not sufficient, as has been mentioned in the note on that de- 

finition i They are therefore now shewn more explicitly. 
' PROP. XXXIV. XXXVI. 

Each of these has a determination, which is now addedj 
which occasions a change in their demonstrations. 

PROP. XXXVII. XXXIX. XL. XLI. 

The 35th and 36th Propositions in the Greek text are 
joined into one, which makes the 39th in this edition, because 
the same enunciation and demonstration serves both : And 
for the same reason Prop, 37. 38. in the Greek are joined 
into one, which here is the 40th. 

Prop. 37. is added to the Data, as it frequently occurs 
in the solution of problems j and Prop. 41. is added, to com- 
plete the rest. 

PROP. XLII. 

This is Prop. 39. in the Greek text, whero the whole con- 
struction of Prop. 22. of Bouk I. of the Elements is put, with- 
out need, into the demonstration, but is now only cited. 

PROP. XI.V. 

This is Prop. 42. in the Greek, where the three straight 
lines made use of in the construction are said, but not shewn, 
to be such that any two of them is greater than the third, 
which is now done. i 



E U C L I D ' S D A T A. 461 

PROP. XLVIT. 

This is Prop. 44. in the Greek text ; but the demonstration 
of it is changed into another, wherein the several cases of it are 
shewn, which, though necessary, is not done in the Greelc. 
PROP. XLVIII. 

There are two cases in this proposition, arising from the 
two cases of the third part of Prop. 47. on which the 48th 
depends ; and in the composition these two cases are expli- 
citly given. 

PROP. LII. 

The construction and demonstration of this, which is Prop- 
48. in the Greek, are made something shorter than in that text, 

PROP. uir. 

Prop. 63. in the Greek text is omitted, being only a case 
of Prop. 49. in that text, which is Prop. 53. in this edition. 

PROP. LVIII. 

This is not in the Greek text, but its demonstration is con' 
tained in that of the first part of Prop. 54. in that text ; which 
proposition is concerning figures that are given in species: This 
58th is true of similar figures, though they be not given in 
species, and, as it frequently occurs, it was necessary to add itj 

PROP. LIX. LXI. 

This is the 54th in the Greek ; and the 77th in the Greek, 
being the very same with it, is left out, and a shorter demon- 
stration is given of Prop. 6i. 

PROP. LXII. 

This, which is most frequently useful, is not in the Greek^ 
and is necessary to Prop. 87. 88. in this edition, as also, though 
not mentioned, to Prop. 86. 87. in the former editions. Prop. 
66. in the Greek text is made a corollary to it. 

PROP. LXIV. 

This contains both Prop. 74, and 73. in the Greek text; 
the first case of the 74th is a repetition of Prop. 56. from 
which it is separated in that text by many propositions ; and 
as there is no order in these propositions, as they stand in the 
Greek, they are now put into the ordei which seemed most 
c<Ml?enient and natural. 

The 



462 N O T E S O N 

The demonstration of the first part of Prop. 73. in the Greek 
is grossly vitiated. Dr. Gregory says, that the sentences he has 
inclosed betwixt two stars are superfluous, and ought to be can- 
celled ; but he has not observed, that what follows them is ab- 
surd, being to prove that the ratio [See his figure] of AT to 
FK is given, which, by th- hypothesis at the beginning of the 
proposicion, is expressly given j so that the whole of this part 
was to be altered, which is done in this Prop. 64. 

PROP. LXVII. LXVIII. 

Prop. 70, in the Greek text, is divided into these two, for 
the sake of distindtness ; and the demonstration of the 67th is 
rendered shorter than that of the first part of Prop. 70. in the 
Greek, by means of Prop. 23. of Book 6. of the Elements. 

PROP. LXX. 

This is Prop. 62. in the Greek text ; Prop. 78. in that text 
is only a particular case of it, and is therefore omitted. 

Dr. Gregory, in the demonstration of Prop. 62. cites the 
49th Prop. dat. to prove that the ratio of the figure AEB to 
the parallelogram AH is given; whereas this was shewn a few 
lines before : And besides, the 49th Prop, is not applicable to 
these two figures ; because AH is not given in species, but is, 
by the step for which the citation is brought, proved to be 
given in species. 

PROP. LXXill. 

Prop. 83. in the Greek text, is neither well eniinciated nor 
demonstrated. The 73d, which in this edition is put in place of 
it, is really the same, as will appear by considering [See Dr. 
Gregory's edition], that A, B, F, E, in the'Greek text, are four 
proportionals, and that the proposition is to shew, that A, 
which has a given ratio to E, is to F, as B is to a straight line 
to which A has a given ratio; or, by inversion, that F is to A, 
as a straight line to which A has a given ratio, is to B : that is, 
if the proportionals be placed in this order, viz. F, E, A, B, that 
the first F is to A, to which the second E has a given ratio, as. 
a straight line to which the third A has a given ratio is to the 
fourth B; which is the enuncia;tion of this 73d, and was thus 
changed, that it might be made like to that of Prop. 72. in 
this edition, which is the 82d in the Greek te^t : And the de- 

3 monstratioa 



EUCLID'SDATA. 463 

monstration of Prop. 73. is the same with that of Prop. 72. 
only making use of Prop. 23. instead of Prop. 22. of Book 5. 
of the Elements. 

PROP. LXXVII. 

This is put in place of Prop. 79. in the Greek text, which 
is not a datum, but a theorem premised as a lemma to Prop. 80. 
in that text: And Prop. 79. is made Cor. i. to Prop. 77. in * 

this edition. CI. Hardy, in his edition of the Data, takes no- 
tice, that in Prop 80. of the Greek text, the parallel KL in the 
figure of Prop. 77. in this edition, must meet the circumfe- 
rence, but does not demonstrate it, which is done here at the 
end of Cor. 3. Prop. 77. in the construction for finding a tri- 
angle similar to ABC. 

PROP. LXXVIII. 

The demonstration of this, which is Prop. 80. in the Gneek 
is rendered a good deal shorter by help of Prop. 77. 

PROP. LXXIX. LXXX. LXXXI. 

These are added to Euclid's Data, as propositions which 
are often useful in the solution of problems. 

PROP. LXXXII. 

This, which is Prop. 60. in the Greek text, is placed before 
the 83d and 84th, which in the Greek are the 58th and 59th, 
because the demonstration of these two in this edition are dedu- 
ced from that of Prop. 82. from which they naturally follow. 

PROP. LXXXVIII. XC. 

Dr. Gregory, in his preface to Euclid's works, which he 
published at Oxford in 1703, after having told that he had sup- 
plied the defects of the Greek text of the Data in mnumerabic 
places from several manuscripts, and corrected Ci. Hardy's 
translation by Mr. Bernard's, adds, that the 8bth theorem, "or 
*' proposition," seemed to be remarkably vitiated, but which 
could not be restored by help of the manuscripts; thta he 
gives three difterent translations of it in Lat:n, according to 
which he thinks it may oe read ; the two first have no distinct 
meaning, and the third, which he say* is the bebt, tnough it 

cofi:ains 



464 N O T E S O N 

contains a true proposition, which is the 90th in this edition, 
has no connection in the least with the Greek text. And it is 
strange that Dr. Gregory did not observe, that, if Prop. 86. 
was changed into this, the demonstration of the 86th must be 
cancelled, and another put into its place :^ But the truth is, both 
the enunciation and the demonstration of Prop. 86. are quite 
entire and right, only Prop. 87. which is more simple, ought 
to have been placed before it ; and the deficiency which the 
* Doctor justly observes to be in this part of Euclid's Data, and 

which, no doubt, is owing to the carelessness and ignorande of 
the Greek editors, should have been supplied, not by changing 
Prop. 86. which is both entire and necessary, but by adding 
the two propositions, which are the 88th and 90th in this 
edition. 

PROP. XCVIII. C. 

These were communicated tome by two excellent geome- 
ters, the first of them by the Right Honourable the Earl of 
Stanhope, and the other by Dr. Matthew Stewart j to which I 
have added the demonstrations. 

Though the order of the propositions has been in many 
places changed from that in former editions, yet this will be 
of little disadvantage, as the ancient geometers never cite 
the Data, and the moderns very rarely. 



XjLS that part of the composition of a problem which is its 
construction may not be so readily deduced from the ana- 
lysis by beginners, for their sake the following example is 
given J in which the derivation of the several parts of the con- 
struction from the analysis is particularly shewn, that they may 
be assisted to do the like in other problems. 

PROBLEM. 

Having given the magnitude of a parallelogram, the angle 
of which ABC is given, and also the excess of the square of its- 
sides BC above the square of the side AB j to find its sides 
and describe it. 

The analysis of this is the same with the demonstration of 
the 87th Prop, of the Data, and the construction that is given 
of the problem at the end of that proposition is thus derived 
from the analysis. 

Let 



EUCLID'S DATA. 

Let EFG be equal lo the given angle ABC, and because 
in the analysis it is said that the ratio of the rectangle AB, 
BC, to the parallelogram AC is given by the 62d Prop. dat. 
therefore, from a point in FE, the perpendicular EG is drawn 
to FG, as the ratio of FE to EG is the ratio of the rectangle 



465 




rpD c 




¥ (t L O HN 



AH^ BC to the parallelogram AC, by what is shewn at tl-.e end 
of Prop. 62. Next, the magnitude of AC is exhibited by 
making the rectangle EG, GH equal to it ; and the given excess 
of the square of BC above the square of BA, to which excess 
the rectangle CB, BD is equal, is exhibited by the rectangle 
HG, GL: Then, in the analysis, the rectangle AB, BC is said 
to be given, and this is equal to the rectangle FE, GH, because 
the rectangle AB, BC is to the parallelogram AC, as (FE to 
EG, that is, as the rectangle) FE, GH to EG, GH ; and the 
parallelogram AC is equal to the rectangle EG, GH, therefore 
the rectangle AB, BC, is equal to FE, GH : And consequently 
the ratio of the rectangle CB, BD, that is, of the rectangle 
HG, GL, to AB, BC, that is, of the straight line DB to BA, 
is the same with the ratio (of the rectangle GL, GH to FE, 
GH, that is) of the straight line GL to FE, which ratio of DB 
to BA, is the next thing said to be given in the analysis : From 
this it is plain that the square of FE is to the square of GL, as 
the square of BA, which is equal to the rectangle BC, CD, is 
to the square of BD : The ratio of which spaces is the next 
thing said to be given: And from this it follows, that four times 
the square of FE is to the square of GL, as four times the 
rectangle BC, CD is to the square of BD; and, by composition, 
four times the square of FE, together with the square of GL, 
is to the square of GL, as four times the rectangle BC, CD, 
together with the square of BD, is to the square of BD, that 
is (8. 6.) as the square of the straight lines BC, CD taken 
together is to the square of BD, which ratio is the next thing 
said to be given in the analysis : And because four times the; 
square of FE and the square of GL are to be added together 
therefore in the perpendicular EG there be taken KG equal to 

H h FE, 



466 N O T E S O N 

FE, and MG equal to the double of it, because thereby thc- 
squares of MG, GL, that is, joining ML, the square of ML> 
is equal to four t,imes the square of FE, and to the square of 
GL : And because the square of ML is to the square of GL, 
as the square of the straight line made up of BC and CD is to 
the square of ^D, therefore (22. 6.) ML is to LG, as BC 
together with CD is to BD j and, by composition, ML and LG 
together, that is, producing GL to N, so that ML be equal to 
LN, the straight line NG is to GL, as twice BC is to BD ; 
and by taking GO equal to the half of NG, GO is to GL, as 
BC to BD, the ratio of which is said to be given in the analy-^ 
sis : And from this it follows, that the rectangle HG, GO is 
to HG, GL, as the square of BC is to the rectangle CB, BD, 
■which is equal to the rectangle HG, GL ; and therefore the 
square of BC is equal to the rectangle HG, GO; and BC is 
consequently found by taking a mean proportional betwixt HG 
and GO, as is said in the construction : And because it was 
shewn that GO is to GL, as' BC to BD, and that now the 
three first are found, the fourth BD is found by 12. 6. It 
was likewise shewn that LG is to FE, or GK, as DB to BA, 
and the three first are now found, and thereby the fourth BA. 
Make the angle ABC equal to EFG, and complete the paral- 
lelogram of which the sides are AB, BC, and the constfuction 
is finished ; the rest of the composition contains the demon- 
stration. 



XXS-the propositions from the J3th to the 28th may ()e 
thought by beginners to be less useful than the rest, be- 
cause they cannot so readily see how they are to be made use of 
in the solution of problems ; on this account the two following 
problems are added, to shew that they are equally useful with 
the other propositions, and from which it may be easily judged 
that many other problems depend upon these propositions. 

PROBLEM L 

X O find three straight lines such, that the ratio 
of the first to the second is given; and if a given 
straight line be taken from the second, the ratio of 
the remainder to the third is given ; also the rect- 
angle contained by the first and third is given. 



E U C L I D'S D A T A. • 467 

Let AB be the first straight line, CD the second, and EF the 
third : And because the ratio of AB to CD is given, and that 
if a given straight line be taken from CD, the ratio of the 
remainder to EF is given ; therefore* the excess of the first AB * 2i. dat. 
above a given straight line has a given ratio to the third EF : 
Let BH be that given straight line; therefore AH the excess 
or AB above it, has a given ratio to EF j and . lt r 

consequently'* the rectangle BA, AH, has a ^ *? ^ " ^* ^' 

given ratio to the rectangle AB, EF, which 

last rectangle is given by the hypothesis; C ' O D 

thLretorc'= the rectangle BA, AH is given, * «. dau . 

and BH the excessof its sides is given ; where- K F 

fore the sides AB, AH are given'' : And be- * 85. dat 

cause the ratios of AB to CD, and of AH to jC NM^^ Q 
EF are given, CD and EF are ' given. 

The Composition. 

Let the given ratio of KL to KM be that which AB is 
required to have to CD ; and let DG be the given straight line 
which is to be taken from CD, and let the given ratio of KM 
to KN be that which the remainder must have to EF j also let 
the given rectangle NK, KO be that to which the rectangle 
AB, EF is required to be equal : Find the given straight line 
BH which is to be taken from AB, which is done, as plainly 
appears from Prop. 24. dat. by making as KM to KL, so GD 
to H B. To the given straight line BH apply' a re£langle equal • t9. 6. 
to LK, KO exceeding by a square, and let BA, AH be its 
sides : Then is AB the first of the straight lines required to be 
found, and by making as LK to KM, so AB to DC, DC will 
be the second : And lastly, make as KM to KN, so CG to 
EF, and EF is the third. 

For as AB to CD, so is HB to GD, each of these ratios 
being the same with the ratio of LK to KM ; therefore*" AH '19. 5. 
is to CG, as (AB to CD, that is, as) LK to KM; and as 
CG to EF, so is KM to KN ; wherefore, ex aequali, as AH 
to EF, so is LK to KN : And as the rectangle BA, AH to 
the rectangle BA, EF, so is* the rectangle LK, KO to the ■ i.g. 
rectangle KN, KO : And by the construction, the rectangle 
BA, AH is equal to LK, KO: Therefore '' the rectangle AB, » 14. j. 
EF is equal to the given rectangle NK, KO : And AB has to 
CD the given ratio of KL to KM ; and from CD the given 
straight^ine GD being taken, the remainder CG has to EF 
the given ratio of KM to KN. Q. E. D. 

Hh2 



46S 



NOTES ON 



PROB. II. 

JL O find three straight lines such, that the ratio 
of the first to the second is given ; and if a given 
straight line be taken from the second, the ratio of 
the remainder to the third is given ; also the sum 
of the squares of the first and third is given. 

Let AB be the first straight line, BC the second, and BD the 
third: And because the ratio of AB to BC is given, and that 
if a given straight line be taken from BC, the ratio of the 

» 24. dat, remainder to BD is given ; therefore, the excess of the first AB 
above a given straight line, has a given ratio to the third BD: 
Let AE be that given straight line, therefore the remainder EB 
has a given ratio to BD : Let BD be placed at right angles to 

*44.dat. EB, and join DE; then the triangle EBD is^ given in species; 
wherefore the angle BED is given: Lei AE which is given in 
magnitude, be given also in position, as also the point £, and 

« 32. dat. the straight line ED will be given'^ in position: Join AD, and 

^ 47. 1. because the sum of the squares of AB, BD, that is^, the square 
of AD is given, therefore the straight line AD is given m mag- 

« 3*. dat. nitudj; and it is also given'^ in position, because from the given 
point A it is drawn to the straight line ED given in position : 
Therefore the point D, in which the two straight lines AD, 

'28.dat, ED, given in position, cut one another, is given*^ : And the 

«.S3. dat straight line DB, which is at right angles to AB, is givens in 
position, and AB is given in position, therefore*^ the point B 

" 29. dat, is given : And the points A, D are given, wherefore *> the 
straight lines AB, BD arc given : And the ratio of AB to BC 

' 2. dat. is given, and therefore' BC is given. 

The Composition. 

Let the given ratio of FG to GH be that which AB is 
required to have to BC,and let HK be the given straight line 
which is to be taken from BC, and let the ratio which the 





F G H X 

remainder is required to have to BD be the given ratio of HQ 
to LG, and place GL at right angles to FH, and join LF, LH : 

Next, 



E U C L I D'S D A T A. 469 

Next, as HG is to GF, so make HK to AE ; produce AE to 
N, so that AN be the straight line to the square of which tb« 
sum of the squares of AB, BD is required to be equal; and 
make the angle NED equal to the angle GFL; and from the 
centre A, at the distance AN, describe a circle, and let its cir- 
cumference meet LD in D, and draw DB perpendicular to AN, 
and DM making the aitgie BDM equal to the angle GLH. 
Lastly, produce BM to C, so that AlC be equal to KH ; then is 
AB the ftrst, BC the second, and BD the third of the straight 
lines that were to be found. 

For the triangles EBD, FGL, as also DBM, LGH being 
equiangular, as EB, to BD, so is FG to GL ; and as DB to 
BiVI, so is LG to GH j therefore, cx aequali, as EB to BM, 
so is (FG to GH, and so is) AE to HK or MC ; wherefore'', * 12.5. 
AB is to BC, as AE to HK, that is, as FG co GH, that is, in 
the given ratio : and from the straight line BC taking MC, 
which is equal to the given straight line HK, the remainder 
BM has to BD the given ratio of HG to GL : and the sum 
of the squares of AB, BD is equai^" to the square of AD or "* *' ^ 
AN, which is the given space. Q. E. D. 

I believe it would be in vain to try to deduce the precedlne 
construction from an algebraical solution of the problem. 



£XD OF NOTES TO THE PATA. 



Uh3 



X 



i 



THE 



ELEMENTS 



OF 



PLANE AND SPHERICAL 



TRIGONOMETRY, 



LONDON: 

Priiited for F. Winceave, in the Strand, Successor to Mr. NornsE, 
1806. 



PLANE TRIGONOMETRY 



473 



LEMMA L Fig. i. 



i^ET ABC be a re£tilineal angle, if about the point B as a 
centre, and with any distance BA, a circle be described, 
meeting BA, BC, the straight lines including the angle ABC 
in A, C ; the angle ABC will be to four right angles, as the 
arch AC to the whole circumference. 

Produce AB till it meet the circle again in F, and throu2;h a 
4raw DE perpendicular to AB, meeting the circle in D, E. 

By 33. 6. Elem. the angle ABC is to a right angle ABD, 
as the arch AC to the arch AD ; and quadrupling the conse- 
quents, the angle ABC will be to four right angles, as the 
arch AC to four time^ the arch AD, or to the whole circum- 
ference. 

LExMMA IL Fig. 2. 

_LiET ABC be a plane rectilineal angle as before : About B 
as a centre with any two distances BD, BA; tet two circles 
be described meeting BA, BC, in D, E, A, C ; the arch AC 
will be to the whole circumference of which it is an arch, 
as the arch DE is to the whole circumference of which it is 
an arch. 

By Lemma i. the arch AC is to the whole circumference 
of which it is an arch, as the angle ABC is to four right an- 
gles j and by the same Lemma i. the arch DE is to the whole 
circumference of which it is an arch, as the angle ABC is to 
four right angles ; therefore tiie.arch AC is to the whole cir- 
cumference of which it is an arch, as the arch DE to the whole 
circumference of which it is an arch. 

DEFINITIONS. Fig. 3. 

I. 

J-iET ABC be a plane reflilineal angle ; if about B as a 
centre, with BA any distance, a circle ACF be described, 
meeting BA, BC, in A, C ; the arch AC is called the measure 
of the anirle ABC. 

\ II. 

The circumference of a circle is supposed to be divided into 

360 



474. PLANE TRIGONOMETRY. 

360 equal parts called degrees^ and each degree into 6a 
equal parts called minutes, and f^ach minute into 60 equal 
parts calL^d seconds, &c. And as many degrees, minutes, 
seconds, &c. as are contained in any arch, ot so many de- 
grees, minutes, seconds, &c. is the angle, of which that arch 
is the measure, said to be. 
Cor. Whatever be the radius of the circle of which the mea- 
sure of a given angle is an arch, that arch will contain the 
same number of degrees, minutes, seconds, &c. as is mani- 
fest from Lemma 2. 

III. 

Let AB be produced till it meet the circle again in F, the angle 
CBF, which, together with ABC, is equal to two right 
angles, is called the «S«^^/^»;^«? of the angle ABC. 

IV. 

A straight line CD drawn through C, one of the extremities 
of the arch AC perpendicular upon the diameter passing 
through the other extremity A is called the Sine of the arch 
AC, or of the angle ABC, of which it is the'measure. 

Cor. The Sine of a quadrant, or of a right angle, is equal to 
the radius. 

V. 

The segment DA of the diameter passing through A, one 
extremity of the arch AC between the sine CD, and that 
extremity, is called the Fersed Sine of the arch AC, or an- 
gle ABC. 

VI. 

A straight line AE touching the circle at A, one extremity of 
the arch AC, and meeting the diameter BC passing through 
the other extremity C in E, is called the Tangent of the arch 
AC, or of the angle ABC. 

VII. 

The straight line BE between the centre and the extremity 
of the tangent AE, is called the Secant of the arch AC, or 
angle ABC. 

Cor. to def. 4. 6. 7. the sine, tangent, and secant of any an- 
gle ABC, arc likewise the sine, tangent, and secant of its 
supplement CBF. 

It is manifest from def. 4. that CD is the sine of the angle 
CBF. Let CB be produced till it meet the circle again in Gj 
and it is manifest that AE is the tangent, and BE the se- 
cant, of the angle ABG or EBF, from def. 6. 7. '- 

CoR.I 



PLANE TRIGONOMETRY. 475; 

Cor. to def. 4. 5. 6. 7. The sine, versed sine, tangent, and ^'S- * 
secant, of any arch which is the measure of any given angle 
ABC, is to the sine, versed sine, tangent, and secant, of 
any other arch which is the measure of the same angle, as 
the radius of the first is to the radius of the second. 

Let AC, MN be me^asures of the angle ABC, according to def. 
I. CD the sine, DA the versed sine, AE the tangent, and 
BE the secant of the arch AC, according to def. 4. 5. 6. 7. 
and N O the sine, OM the versed sine, MP the tangent, and 
BP the secant of the arch MN, according to the same defini- 
tions. Since CD, NO, AE, MP are parallel, CD is to NO 
as the radius CB to the radius NB, and AE to MP as AB to 
BM, and BC or BA to BD, as BNor BM to BO ; and, by 
conversion, DA to MO as AB to MB, Hence the corollary 
is manifest ; therefore, if the radius be supposed to be di- 
vided into any given number of equal parts, the sine, versed 
sine, tangent, and secant of any given angle, will each con- 
tain a given number of these parts; and, by trigonometrical 
tables, the length of the sine, versed sine, tangent, and se- 
cant of any angle may be found in parts of which the radius 
contains a given number ; and, vice versa, a number expres- 
sing the length of the sine, versed sine, "tangent, and secant 
being given, the angle of which it is the sine, versed sine, 
tangent, and secant, may be found. 



Vin. Fig. a. 

The difference of an angle from a right angle, is called the 
complement of that angle. Thus, if BH be drawn perpen- 
dicular to AB, the angle CBH will be the compliment of 
the angle ABC, or of CBF. 



IX. 

Let HK be the tangent, CL or DB, which iseenjal to It, th* 
sine, and BK the secant of CBH, the complement "of.^BC, 
according to def. 4. 6. 7. HK is called the cotangent, BD 
the cosine^ and BK the ctsecant of the angle ABC. 

CoR. I, The radius is a mean proportional between the tan- 
gent and cotangent. 

For, since HK, BA are parallel, the angles HKB, ABC will 
be equal, and the angles KHB, BAE are right i therefore 

the 



47^ PLANE TRIGONOMETY. 

the triangles BAE, KHB are similar, and therefore AE is to 
AB, asBHorBAtoHK. 

Cor. 2. The radius is a mean proportional between the co- 
sine and secant of any angle ABC. 

Since CD, AE are parallel, BD is to BC or BA, as BA t* 
BE. 

PROP. I. Fig. 5. 

In a right angled plane triangle, if the hypothe- 
nuse be made radius, the sides become the sines of 
the angles opposite to them ; and if either side, be ^ 
made radius, the remaining side is the tangent of '^ 
the angle opposite to it, and the hypothenuse the 
secant of the same angle. 

Let ABC be a right angled triangle ; if the hypothenuse 
BC be made radius, either of the sides AC will be the sine of 
the angle ABC opposite to it; and if either side BA be made 
radius, the other side AC will be the tangent of the angle ABC 
opposite to it, and the hypothenuse BC the secant of the same 
angle. 

About B as a centre, with BC, BA for distances, let two 
circles CD, EA be described, meeting BA, BC in D, E: Since 
CAB is a right angle, BC being radius, AC is the sine of the 
angle ABC, by def. 4. and BA being radius, AC is the tan- 
gent, and BC the secant of the angle ABC, by def. 6. 7. 

Cor. I. Of the hypothenuse a side and an angle of a right 
angled triangle, any two being given, the third is also given. 

Cor. 2. Of the two sides and an angle of a right angled 
triangle, any two being given, the third is also given. 



PROP. n. Fig. 6. 7.^ 

1 HE sides of a plane triangle are to one another, 
as the sines of the angles opposite to them. 

hi right angled triangles, this Prop, is manifest from Prop. i. 
for if the hypothenuse be made radius, the sides are the sines of 
the angles opposite to them, and the radius is the sine of aright 
angle (cor. to def. 4.), which is opposite to the hypothenuse. 

In 



PLANE TRIGONOMETRY. 

In any oblique angled triangle ABC, any two sides AB, 
AC will be to one another as the sines of the angles ACB, 
ABC, which are opposite to them. 

From C, B draw CE, BD perpendicular upon the opposite 
sides AB, AC produced, if need be. Since CEB, CE)B are 
rightangles, BC being radius, CE is the sine of the angle CB A, 
and BD the sine of the angle ACB ; but the two triangles 
CAE, DAB have each a right angle at D and E j and like- 
wise the common angle CAB ; thereKire they are similar, and 
consequently, CA is to AB, as CE to DB ; that is, the sides 
are as the sines of the angles opposite to them. 

Cor. Henre ot two sides, and two angles opposite to them, 
in a plain triangle, any three being given, the fourth is also 
given. 

PROP. III. Fig. 8. 

-i-N a plane triangle, thesum of any two sides is to 
their difference, as the tangent of half the sum of 
the angles at the hase, to the tangent of half their 
difference. 

Let ABC be a plane triangle, the sum of any two sides 

AB, AC will be to their difference as the tangent of half the 
sum of the angles at the base ABC, ACB to the tangent of 
half their difference. 

About A as a centre, with AB the greater side for a distance 
let a circle be described, meeting AC produced in E, F, and 
BC in D ; join DA, EB, FB : and draw FG parallel to BC, 
meeting EB in G. 

The angle EAB (32. i.) is equal to the sum of the an^rles 
at the base, and the angle EFB at the circumference is equal 
to the half of EAB at the centre (20. 3.) ; therefore EFB is 
half the sum of the angles at the base ; but the antrle ACB 
(32. I.) is equal to the angles CAD and ADC, or ABC to- 
gether ; therefore FAD is the difference of the angles at the 
base, and FBD at the circumference, or BFG, on account of 
the parallels FG, BD, is the half of that differtnce; but since 
the angle EBF in a semicircle is a right angle (i. of this), 
FB being radius, BE, BG are the tangents of the angles EFB, 
BFG ; but it is manifest that EC is the sum of the sides £A, 

AC, and CF their difference ; and since BC, FG are parallel 
(2 6.) EC is to CFj as EB to BG ; that is, the sum of the 

sides 



477 



478 PLANE TRIGONOMETRY. 

sides is to their difference, as the tangent of half the sum of 
the angles at the base to the tangent of half their difference'. 

PROP. IV. Fic. 18. 

xN any plane triangle BAG, whose two sides arc 
BA, AC, and base BC, the less of the two sides 
which let be BA, is to the greater AC as the radius 
is to the tangent of an angle, and the radius is to 
the tangent of the excess of this angle above half a 
right angle as the tangent of half the sum of the 
angles B and C at the base, is to the tangent of 
half their difference. 

At the point A draw the straight line EAD perpendicular 
to BA J make AE, AF, each equal to AB, and AD to AC ; 
join BE, BF, BD, and from D, draw DG perpendicular upon 
BF. And because BA is at right angles to EP\ and EA, AB, 
AF are equal, e«ch of the angles EBA, ABF is half a right 
angle, and the whole EBF is a right angle (also 4. i. El.) j 
EB is equal to BF. And since EBF, FGD are right angles, 
EB is parallel to GD, and the triangles EBF\ FGl3 are simi- 
lar i therefore EB is to BF, as DG to GF, and EB being 
equal to BF, FG must be equal to GD. And because BAD 
is a right angle, BA the less side is to AD or AC the great- 
er as the radius is to the tangent of the angle ABDj 
and because BGD is a right angle, BG is to GD or GF as 
the radius is to the tangent of GBD, which is the excess of 
thea;ngle ABD above ABF half a right angle. But because 
EB is parallel to GD, BG is to GF as ED is to DF, that is, 
since ED is the sum of the sides BA, AC, and FD their dif- 
ference (3. of this), as the tangent of half the sum of the an- 
gles B, C, at the base to the tangent of half their difference. 
Therefore, m any plane triangle, &c. Q. E. D. 

PROP. V. Fig. 9. and lo. 

J-N any triangle, twice the rectangle contained by 
any two sides is to the difference of the sum of the 
squares of these two sides, and the square of the 
base, as the radius is to the cosine of the angle in- 
cluded by the two sides. 

Let ABC be a plane triangle, twice the rectangle ABD con- 
tained by any two sides BA, BC is to the difference of the su^ 

of 



PLANE TRIGONOMETRY. 475 

of the squares of BA, BC, and the square of the base AC, as 
the radius to the cosine of the angle ABC. 

From A, draw AD perpendicular upon the opposite side 
BC, then (by 12. and 13. 2. El.) the difference ot the sum of 
the squaies of AB, BC, and the square of the base AC, is 
er^ual to twice the rectangle CBD ; but twice the rectangle 
CBA is to twice the rectangle CBD ; that is, to the differ- 
ence of the sum of the squares of AB, BC, and the square of 
AC (i. 6.) as AB to BD ; that is, by Prop. i. as radius to 
the sine ot BAD, which is the complement of the angle ABC, 
that is, as radius to the cosine of ABC. 

PROP. VI. Fig. IJ. 

JlN any triangle ABC, whose two sidesare AB, AC, 

and base BC, the rectangle contained liy half the pe- 
rimeter, and the excess of it above the base BC, is to 
the rectangle contained by the straiglit lines by 
which the half of the perimeter exceeds, the other 
two sides AB, AC, as the square of the radius is to 
the square of the tangent of half the angle BAC 
opposite to the base. 

Let the angles BAC, ABC be bisected by the straight lines 
AG, BG i a. id producing the side AB, let the exterior angle 
CBH be bisected by the straight line BK, meeting AG in K; 
and from the points G, K, let there be drawn perpendicular 
upon the sides the straight lines GD, GE, GF, KH, KL, 
KM, Since therefore (4, 4.) G is the centre « the circle in- 
scribed in the triangle ABC, GD, GF, GE will be equal, and 
AD will be equal to AE, BD to BF, and CE to CF. In like 
manner KH, KL, KM Will be equal, and BH will be equal 
to.BM and AH to AL, because the angles HBM, HAL are 
bisected by the straight lines BK, KA : And because in the 
triangles KCL, KCM, the sides LK, KM are equal, KC is 
common, and KLC, KMCare right angles, CL will be equal 
to CM: Since therefore BM is equal to BH,and CM to C'L; 
BC will be e4ual to BH and CL together j and, adding AB 
ajid AC together, AB, AC, and BC will together be equal 
to AH and AL together : But AH, AL are equal : Wherefore 
each of them is equal to half the perimeter of the triangle 
ABC : But since AD, AE are equal, and BD, BF, and also 
CE, CF, AB, together with FC, will be equal to half the 
perimeter of the triangle to which AH or AL was. shewn to be 
equal i taking away therefore ibe common AB, the remainder 

FC 



48o PLANE TRIGONOMETRY. 

FC will be equal to the remainder BH : In the same manner 
it is demonstrated, that BF is equal to CL : And since the 
points B, D, G, F, are in a circle, the angle DGF will be 
equal to th- exterior and opposite angle FBH (22, 3.) j where- 
fore their halves BGD, IIBK will be equal to one another : 
The right angled triangles BGD, HBK will therefore be 
equiangular, and GD will be to BD, as BH to HK, and tho 
redangle contained byGD, HK will be equal to the rectangle 
DBH or BFC : But since AH is to HK, as AD to DG, the 
rc£langle HAD (22. 6.) will be to the rectangle contained by 
HK, DG, or the redtangle BFC (as the square of AD is to 
the square of DG, that is) as the square oi the radius t'o the 
square of the tangent of the angle DAG, that is, the half of 
33 AC : But HA is half the perinieter of the triangle ABC, and 
AD is the excess of the same above HD, that is, above the 
base BC ; but BF or CL is the excess of HA or AI. above the 
side AC, and FC, or HB, is the excess of the same HA above 
the side AB ; therefore the rectangle contained by half the 
perimeter, and the excess of the same above the base, viz. the 
redangle HAD, is to the redtangle contained by the straight 
• lines by which the half of the perimeter exceeds the other two 
side, that is, the redtangle BFC, as the square of the radius is 
to the square of the tangent of half the angle BAC opposite to 
the base. Q. E. D. 

PROP. VH. Fig. 12. 13. 

In a plane triangle, the base is to the sum of the 
sides as the difference of the sides is to the sum or 
difference of the segments of the base made by the 
perpendicular upon it from the vertex, according as 
the square of the greater side is greater or less than 
the sum of the squares of the lesser side and the base. 

Let ABC be a plane triangle ; if from A the vertex be drawn 
a straight line AD, perpendicular upon the base BC, the base 
BC will be to the sum of the sides BA, AC, as the difference 
of the same sides is to the sum or difference of the segments 
CD, BD, according as the square of AC the greater side is 
- greater or less than the sum of the squares of the lesser side 
AB, and the base BC. 

About A as a centre, with AC the greater side for a 
distance, let a circle be described meeting AB produced in K, 
F, and CB in G : It is manifest, that FB is the sum, and BE 

the 



PLANE TRIGOl^OMETRY. 481 

the difference of the sides ; and since AD is perpendicular to 
GC, GD, CD will be equal ; consequently GB will be equal 
to the sum or difference of the segments CD, BD, according 
as the perpendicular AD meets the base produced, or the base ; 
that is (by Conv. 12. 13. 2.), according as the square of AC is 
greater or less than the sum of the squares of AB, BC : But 
(by 55. 3.) the redangle CBG is equal to the redangle EBF; 
that is ( 16. 6. ) BC is to BF, as BE is to BG ; that is, the base 
.is to the sum of the sides, as the difference of the sides is to the 
sum or difference of the segments of the base made by the per- 
pendicular from the vertex, accordmg as the square of the 
greater side is greater or less than the stem of the squares of 
the lesser side and the base. Q. E. D. 

PROP. VIII. PROB. Fig, 14. 

Jl HE sum and difference of two magnitudes being 
given, to find them. 

Half the given sum added to half the given difference, will 
be the greater, and half the difference snbtracted fromlialf the 
sum, will be the less. 

For let AB be the given sum, AC the greater, and BC the 
less. Let AD b? half the given sum ; and to AD, DB, which 
are equal, let DC be added ; then AC will be equal to BD, 
and DC together ; that is, to BC, and twice DC ; conse- 
quently twice DC is the difference, and DC half tliat diffe- 
rence ; but AC the greater is equal so AD, DC ; that is, to 
half the sum added to half the difference, and BC the less is 
equal to the excess ot BD, half the sum, above DC half the 
difference. Q^ E. D. 

SCHOLIUM. 

Of the six parts of a plain triangle (the three sides and three 
angles) any three being given, to find the other three is the 
business of plane trigonometry ; and the several cases of that 
problem may be resolved by means of the preceding proposi- 
tions, as in the two following, with the tables annexed. In 
these, the solution is expressed by a fourth proportional .to 
three given lines ; but if the given parts be expressed by num- 
bers from trigonometrical tables, it may be obtained arithme- 
tically by the common Rule of Three. 



NoTF. In the tables the following abbreviations are used: R. is put 
for the Radius; T. for Tangent; and S, for Sine. Degrees, minutes, 
seconds, &c. are written in this manner : 30* 25' IS", &c, which signiiie* 
ao degrees, 25 minutes, 13 seconds, &c. 

I i 



482 



PLANE TRIGONOMETRY. 



SOLUTION of the Cases of Right Angled 
Triangles. 



Fig. 15. 



GENERAL PROPOSITION. 

XN a liglit angled triangle, of the three sides and 
three angles, any two heing given besides the right 
angle, the other three may be found, except when 
the two acute .angles are given ; in which case the 
ratios of the sides are only given, being the same 
with the ratios of the sines of the angles opposite 
to them. 

It is manifest from 47. i. that of the two sides and hypo- 
thenuse, if any two be given, the third may also be found. It 
is also manifest from 32.. i. that if one of the acute angles of a 
right angled triangle be given, the other is also given, for it 
is the complement of the former to a right angle. 

If two angles of any triangle be given, the third is also 
given, being the supplement of the two given angles to two 
right angles. 

The other cases may be resolved by help of the preceding 
propositions, as in the following table : 



G 



iven. 



Sought. 



Two sides, AB,! Theangles 



AC. 



iB, C. 



AB,BC,asideand| Theangles 
the hypothenuse. IB, C. 



AB, B, a side and The other 
an angle. side AC. 



AB and B, a side 
and an angle. 



BC and B, the 
hypothenuse and an 
angle. 



The hypo- 
thenuse BC. 



The side 
AC. 



AB:AC::R:T,B,of 

wh ich C is the complement 



Be : BA : : R : S, C, of 

which B is the complement 



R:T,B::BA: AC. 



S, C : R : : BA : BC. 



R : S, B : : BC : CA. 



These five cases are resolved by Prop. i. 



PLANE TRIGONOMETRY 



483 



SOLUTION ofthe Cases of Oblique Anqled 
Triangles. 



GENERAL PROPOSITION. 

XN an oblique angled triangle, of the three sides 
and three angles, any three being given, the other 
three may be found, except when the three angles 
are given ; in which case the ratios of the sides are 
only given, being the same with the ratios of the 
sines of the angles opposite to them. 
Given. Sought. 



A, B, and there- BC, AC, 
fore C, and the side 
AB. 


S, C : S, A : : AB : BC, 
and also S, C : S, B : : AB 
:AC.(2.) 


AB, AC, and B, 
two sides and an 
angle opposite to 
one of them. 


The angles 
A and C. 


AC : AB : : S, B, S, C, 

(2.) This case admits of 
two solutions j for C may 
be greater or less than a 
quadrant. (Cor. todef. 4.) 


AB, AC, and A, 
two sides, and the 
included angle. 


The angles 
B and C. 


AB + AC:AB-AC::T, 
C + B T, C— B. 

{ J ) 




\S ) 

the sum and difference of 
the angles C, B, being 
given, each of them is 
given. (7.) Otherwise 
Fig. 18. 

BA:AC::R:T,ABD, 
and also R : T,ABD— 45= 

B + C B— C. 
:T,— .T,-- :(4.) 

therefore B and C are 
^ivenas before, (7.) 



Fio. 16. 17 



1 2 



484 



PLANE TRIGONOMETRY. 



Given, 



Sought. 



AB, BC, CA, 

the three sides 



A,B,C,the 
three angles 



2ACxCB;ACq+CBq 

ABq : : R : Co S, C. If 
ABq -f CBq be greater than 
ABq. Fig. 16. 

2ACxCB:ABq— ACq 
-CBq::R:Co. S,C. If 
V Bq be greater than ACq + 
CBq. Fig. i7. (4.) 
Otherwise 

Let AB + B C +AC-2 P. 
P X P — AB : P — AC X 
P — BC : : Rq : Tq, \ C, 



and hence C is known 
Otherwise 

Let AD be perpendicular 
toBC. I. If ABq be less 
than ACq + CBq. FiG. 16. 
BC : BA + AC : : BA 
AC : BD — DC, and BC 
the sum of BD, DC is given ; 
therefore each of them is 
given. (7.) 

2. If A Bq be greater than 
ACq+CBq. F1G.17.BC: 
BA + AC: :BA— AC:BD 
+ DC ; and BC the difltc- 
rence of BD, DC is given, 
therefore each of them is 
given. (7.) 

And CA : CD : : R : Co 
S, C. (i.) and C being 
found, A and B are found 
by case 2. or 3 



HI G O N O 31 ETJR Y. 




lA B 



I) C 



^X/ 



Ti 



^%-^7- 



—JB 



CONSTRUCTIONS 4«5. 

OF THE 

TRIGONOMETRICAL CANON. 



A Trigonometrical Canon is a Table, which, beginning from 
one second or one minute, orderly expresses the lengths 
that every sine, tangent, and secant have, in respect of the 
radius, which is supposed unity ; and is conceived to be 
divided into looooooo or more decimal parts. And so 
the sine, tangent, or secant of an arc, may be had by help 
of this table ; and, contrariwise, a sine, tangent, or secant, 
being given, we may find the arc it expresses. Take no- 
tice, that in the following tract, R signifies the radius, S a 
sine, Cos. a Cosine, T a tangent, and Cot. a cotangent j 
also AC q signifies the square of the right line AC ; and 
the marks or characters +, — , zr, :, : :, and v'j are, seve- 
rally, used to signify addition, subtraction, equality, pro- 
portionality, and the extraction of the square root. Again, 
when a line is drawn over the^um or difference of two 
quantities, then that sum or difference is to be considered as 
one quantity. 



Constructions of the Trigonometrical Can^n. 
PROP. I. THEOR. 

1 HE two sides of any right angled triangle be- 
ing given, the other side is also given. 

For (by 47. I.) ACq=ABq + BCqand ACq-BCq= ri£.98. 
AB q and interchangeably AC q—AB qizBCq. Whence, 
by the extraction of the square root, there i s given AC = 
• ABq + BC q; and AB= -/ ACq-BC qi and BC= 
VACq-ABq. 

PROP. II. PROB. 

1 HE sine DE of the arc BD, and the radius CD Fig. at. 
being given, to find the cosine DF. 

Ii3 Th« 



486 CONSTRUCTIONS OF THE 

The radius CD, and the sine DE, being given in the right 
a ngled triangle CPE, there will be given (by the last Prop.) 
'v/CDq-DEq=(CE = ) DF. 

PROP. III. PROB. 

'^^ ■ 1 HE sine DE of any arc DB being given, to 
find DM or BM, the sine of half the arc. 

DE being given, CE (by the last Prop.) will be given, and 
accordingly EB, which is the difference between the cosine 
and radius. Therefore DE, EB, being given, in the right 
angled triangle DBE, there will be given DB, whose half 
DM is the sine of the arc DLrrf the arc BD. 

PROP. IV. PROB. 

- 1 HE sine BM of the arc BL being given, to find 
th^ sine, of double, that arc. 

?ig. 29. The sine BM being given, there will be given (by Prop. 
2.) the cosine CM. But the triangles CBM, DBE, are equi- 
angular, because the angles at E and M are right angles, and 
the angle at B common : Wherefore (by 4. 6.) we have 
CB : CM : : (BD, or) 2 BM : DE. Whence, since the 
. three first terms of this analogy are given, the fourth also, 
which is the sineof thearc DB,will be known. 

Cor. Hence CB : 2 CM ; : BD : 2 DE j that is, the ra- 
dius is to double the cosine of one half of the arc DB, as the 
subtense of the arc DB is to the subtense of double that arc. 
^AlsoCB: 2CM: :(2BM:2DE: :)BM:DE : :iCB : 
* CM. Wherefore tlie sind of an arc, and the sine of its dou- 
ble, being given,' the cosine of the arc itself is given. 

PROP. V. PROB. 

Pig. 30. j^ pj£ sines of two arcs, BD, FD, being given, to 
find FT, the sine of the sum, as Hkewise EL, the 
sine of their difference. 

Let the radius CD be drawn, and then CO is the cosine of 
the arcFD, which accordingly is given, and draw OP through 
O parallel to DK ; al^o let OM, GE, be ^rawn parallel to 
CB: Then because the triangles CDK, COP, CHI, FOH, 

FOP4 



TRIGONOMETRICAL CANON. 4S7 

FOM, are equiangular ; \a the first place CD : DK : : CO : 
OP, which, consequently, is known. Also we have CD : 
CK : : FO : FM j and so, likewise, this will be known. But 
because F0=EO, then will FM=MG=ON ; and so OP 
+ FM=:FI— sine of the sum of the arcs: And OP— FM; 
ihat is, OP— ON=EL=:sine of the difference of the arcs; 
which were to be found. 

Cor. Because the differences of the arcs BE, BD, BF, are 
equa , the arc BD is an arithmetical mean Between the arcs 
BE, BF. 

PROP. VI. THEOR. 

JL HE same Things being supposed, the radius is 
to double the cosine of the mean arc. as the sine 
of the difference is to the difference of the sines of 
the extremes. 

For we have CD : CK : : FO : FM ; whence, by doubling Fig. 3f. 
the consequents, CD : 2 CK : : FO : (2 FM, or) to FG, 
which is the difference of the sines, EL, FI. Q^ E. D. 

Cor. If the arc BD be 60 degrees, the difference of the 
sines FI, EL, will be equal to the sine FO, of the difference. 
For, in this case, CK is the sine of 30 degrees ; the double 
whereof is equal to the radius (by 15. 4.) ; and so, «inceCD 
=.2 CK, we shall have FO=:FG. And, consequently if 
the two arcs BE, BF, are equidistant from the arc of 60 de- 
grees, the difference of the sines will be equal to the sine of 
the difference FD. 

Cor. 2. Hence, if the sines of all arcs distant from one 
another by a given interval, be given, from the beginning of 
a quadrant to 60 degrees, the other sines may be found by one 
addition only. For the sine of 61 degrees =the sine of 59 
degrees + the sine of i degree; and the sine of 62 degreesz: 
the sine of 58 degrees + the sine of 2 degrees. Also, the sine 
of 53 Degreesrzthe sine of 57 degrees + the sine of 3 degrees, 
and so on. 

Cor. 3. If the sines of all arcs, from the beginning of a 
quadrant to any part of a quadrant, distant from each other 
by a given interval, be given, thence we may find the sines 
of all arcs to the double of that part. For example, let all 
the sines to 15 degrees be given ; then, by the preceding ana- 
logy, all the smes to 30 degrees may be found. For the ra- 
dius is to double the cosine of 15 degrees, as the sine of i de- 

Xi4 grce 



488 CONSTRUCTIONS OF THE 

gree is to the difference of thesines of 14 degrees, and i6 de- 
grees : So, also, is the sine of 3 degrees to the difference be- 
tween the sines of 12 and 18 degrees ; and so on continually, 
until you come to the sine of 30 degrees. 

After the same manner, as the radius is to double the cosine 
of 30 degrees, or to double the sine of 60 degrees, so is the 
sine of i degree to the difference of the sines of 29 and 31 de- 
grees : : sine 2 degrees to the' difference of the sines of 28 
and 32 degrees : : sine 3 degrees, to the difference of the 
sines of 27 and 33 degrees. But, in this case, the radius is 
to double the cosine of 30 degrees, as i to -v/ 3. 

For (see the figure for Prop. 15, Book IV. of the Ele- 
ments) the angle BGC— 60 degrees, as the arc BC, its mea- 
sure, is a sixth part of the whole circumference ; and the 
straight line BC=:R. Hence it is evident that the sine of 30 
degrees is equal to half the radius ; and therefore, by Prop. 2. 

the cosine of 30 degrees =: v'R*— _r::-^_A-_> and its double 

~v'3R*rr:R X V 3. Consequently, radius is to double 
the cosine of 30" : : R : R X a/3 : : i : \/ 3- 

And, accordingly, if the sines of the distances from the arc 
of 30 degrees, be multiplied by -/ 3, the differences of the 
sines will be had. 

So, likewise, may the sines of the minutes in the beginning 
of the quadrant be found, by having the sine, and cosines of 
one and two minutes given. For, as the radius is to double 
the cosine of 2' ; : sine i' ; difference of the sines of I'and 3' : : 
sine 2 : difference of the sines of o' and 4' ; that is, to the 
sine of 4'. And so, the sines of the four first minutes being 
given, we may thereby find the sines of the others to 8' and 
from thence to 16', and so on. 



PROP. VII. THEOR.' 

In smalt arcs, the sines and tangents of the same 
arcs are nearly to one another, in a ratio of equa- 
lity. 

^ig.31. Pqj.^ betause the Triangles CED, CBG, are equiangular, 

CE : CB : : ED : BG. But as the point -E approaches B, 

EB will vanish in respect of the arc BD ; whence CE 

2 will 



TRIGONOMETRICAL CANON. 489 

will become nearly equal to CB, and so ED will be also 

nearly equal to BG. If EB be less than the 

^ ^ I 0000000 

part of the radius, then the difference between the sine and the 

, . I 

I tansient will be also less than the part of the 

• ^ looooeoo ^ 

tangent. 

CoR. Since any arc is less than the tangent, and greater 
than its sine, and the sine and tangent of a very small arc are 
nearly equal ; it follows, that the arc will be nearly equal to 
its sine ; And so, in very small arcs, it will be, as arc is to 
arc, so is sine to sine. 



To 



PRO?. VIII. PROB. 
find the sine of the arc of one minute. 



The side of a hexagon inscribed In a circle, that is, the 
subtense of 60 degrees, is equal to the radius (by Coroll. 15th 
of the 4.th) ; and so the half of the radius will be the sine of 
the arc of 30 degrees. Wherefore the sine of the arc of 30 
degrees being given, the sine of the arc of 15 degrees may be 
found (by Prop. 3.) Also the sine of the arc of 15 degrees 
being given (by the same Prop.)* we may have the sine of 7 
degrees 30 minutes. So, likewise, can we find the sine of 
the half of this, viz. 3 degrees 45 minutes; and so on, until 
12 bisections being made, we come to an arc of 52% 44', 03*, 
45% whose cosine is nearly equal to the radius ; in which 
case (as is manifest from Prop. 7.) Arcs are proportional to 
their sines : and so, as the arc of 52% 44% 03*, 45', is to an 
arc of one minute, so will the sine before found be to the sine 
of an arc of one minate, which therefore will be given. And 
when the sine of one minute is found, then (by Prop, 2. and 
4.) the sine and cosine of two minutes will be had. 

PROP. IX. THEOR, 

If the angle BAC, being in the peripher}- of a 
.circle, be bisected by the right hne AJD, and if AC 
be produced until DE=:AD meets it in E ; then 
willCEzzAB. 

In the quadrilateral figure ABDC (by 2a. 3.) the angles B 

and 



fig. a«. 



490 CONSTRUCTIONS OF THE 

and DC A are equal to two right anglesrrDCE + DCA (by 
13. I.) : whence the angle BrrDCE. But, likewise, thean- 
. gle E=DAC (by 5. i.)=:DAB, and DCrzDB : Wherefore 
the triangles BAD and CED are congruous, and so C£ is 
equal to AB. Q, E. D. 



Fig. S3. 



PROP. X. THEOR. 

Let the arcs AB, BC, CD, DE, EF, &c. be 
eq^^al ; and let the subtenses of the arcs AB, AC, 
AD, AE, &c. be drawn ; then will AB : AC : : AC 
:AB + AD::AD:AC + AE::AE:AD + AF::AF 
:AE + AG. 

Let AD be produced to H, AE to I, AF to K, and AG to 
L, ,so that the triangles ACH, ADI, AEK, AFL, be 
isosceles ones : Then, because the angle BAD is bisected, 
we shall have DHzrAB (by the last Prop.) j so likewise EI 
=:AC, FK=AD, also GL:=AE. 

But the isosceles triangles ABC, ACH, ADI, AEK, AFL, 
because of the equal angles at the bases are equiangular : 
Wherefore it will be, as AB : AC : : AC : (APi=) AB + 
AD : : AD : (Air:) AC + AE : : AE : (AKzr) AD + AF 
:: AF: (AL = ) AE + AG. Q. E. D. 

CoR. I. Because AB is to AC, as radius is to <louble the 
cosine of | the arc AB, (by Coroll. Prop. 4.) it will also be 
as radius is to double the cosine of | the arc AB, so is 4 AB 
: i AC : : ^ AC : I AB + i AD : : ^ AD : \ AC + i AE : : 
I AE : i AD + i AF,&c. Nowlet each of the arcs AB, BC, 
CD, &c. be 2' ; then will \ AB be the sine of one minute, \ AC 
the sine of 2 minutes, f AD the sine of 3 minutes, \ AE the 
sine of 4 minutes, &c. Whence, if the sines of one and two 
minutes be given, we may easily find all the other sines \vt the 
following manner. ■ , - 

Let the cosine of the arc of one minute, that is, the sine of 
the arc of 89 deg. 59' be called Q_; and make the following 
analogies ; R.:%Qj.: Sin. 2' : S. i' +S. 3'. Wherefore thai 
sine of 3 minutes will be given. Also, R. : 2 Q^: : S. 3' :! 
S. 2' +S. 4'. Wherefore the S. 4' is given. And R. : 2 Qj 
: : S. 4' : S. 3' -j-S. 5' j and so the sine of 5' will be had. 

Likewise, R. : 2 Q^: : S. 5' : S 4' +S. 6' ; and so we shall 
have the sine of 6'. And in like manner, the sines of every 
minute of the quadrant will be given. And because the ra- 
dius, or the first term of the analogy, is unity, the operations 

wiU 



TRIGONOMETRICAL CANON. 

will be with great ease and expedition calculated by multipli- 
cation, and contracted by addition. When the sines are 
found to 60 degrees, all the other sines may be had by addi- 
tion only, by Cor. i. Prop. 6. , 

The sines being given, the tangents and secants may be 
found from the following analogies (see Figure 3, for the de- 
finitions) J because the triangles BDC, BAE, BHK are equi- 
angular, we have 

BD : DC : : BA : AE ; that is, Cos. : S. : : R. : T. 

AE : BA : : BH : HK ; that is, T. : R. : : R. : Cot. 

BD : BC : : BA : BE J that is, Cos. : R. : : R. : Secant, 

CD : BC : : BH : BKi that is, S. : R. : : R. : Cosec. 



49 » 



492 OF LOGARITHMS. 



I. X HE indices or exponents of a series of numbers in geo- 
metrical progression, proceeding from i, are also called the 
logarithms of the numbers in that series.* Thus if <2 denote 
any number, and th^ geometrical scries, i, «*, c% a^^ a\ Sec. 
be produced by actual multiplication, then i, 2, 3, 4, &5. are 
called the logarithms of the first, sepond, third, and fourth 
powers of a respectively. Consequently, if, in the above, a 
be equal to the number 2, then i is the logarithm of 2, 2 is 
the logarithm of 4, 3 is the logarithm of 8, 4 is the logarithm 
of 16, &CC. But if fl be equal to lO, then i is the logarithm 
of 10, 2 is the logarithm of 100, 3 is the logarithm of lOOO, 
4 is the logarithm of lOOOO, &c. The series may be conti- 
nued both ways from i. Thus — , -> —1 — » i, a, a\ a' 

/2*, &c. constitute a series in geometrical progression, and, 
agreeable to the established notation in algebra, the indices, 
or logarithms are —4, — 3, — 2, — 1,0, 1,2, 3, 4, &c. If^be 

equal to the number 2, then— 4 is the logarithm of ->, — 3 is 

the logarithm of— , — 2 is the logarithmof — , — i is the loga- 
j 8 4 

rithmof — , o is the logarithmof i, i is the logarithm of2,&c. 

If a be equal to 10, then —4 is the logarithm of , —3 

^ ' ^ ° loooo ^ 

is the logarithm of , — 2 is the logarithm of — , — i is the 

^ , 1000 100 

logarithm of — , o is the logarithm of i, and i is the loga- 

rithm of 10, &c. 

2. From the above it is evident, that the logarithms of a 
series of numbers in geometrical progression, constitute a 
series of numbers in arithmetical progression. Beginning 
with I, and proceeding towards the right hand, the terms in 
the geometrical series are produced by multiplication, but 
their corresponding logarithms are produced by addition. — 
On the contrary, beginning with i, and proceeding towards 
the left hand, the terms in the geometrical progression are 
produced by division, but their corresponding logarithms are 
produced by subtraction. 

The 

• The reader ought to be acquainted with arithmetical and geometrical progressioai 
and the binomial theorem, before he enters on a perusal of any account of logarithms. 



OF LOGARITHMS. 4^j 

3. The same observations apply to logarithms when they • 
:e fractions. Thus if a" denote any number, -^ — , — , 

L 1 L ± aT. a~. a-a . 

p I, «", a", ««j a", &c. constitute a series of numbers ' 

geometrical progression, of which — — , — — , — ^ •—- ^> 
-^ — , i, -t, &c. are the logarithms ; and it is evident that 

e assertions in the last article hold true both with respect 
the numbers in geometrical progression and their corres- 
iding logarithms. As a and n may be taken at pleasure, 
follows that numbers in very different geometrical pro- 
essions may have the same logarithms ; and that the same 
ies of numbers in geometrical progression may have diiFe- 
it scries of logarithms corresponding to them. 

4. If a be an indefinitely small decimal fraction, and suc- 
sive powers of i -f ^ be raised, then the excess of any power 
r 4- ^ above that immediately preceding it will be indefi- 
ely small. Thus let /z = -00000000001, and then i+V^z: 
JOOOO00000200000000001 ; and i + a^'m -00000000003000- 
30000300000000001 ; and proceeding by actual multiplica- 
n to obtain higher powers of 1 -0000000000 1, it will be 
md that the dift'erence between two successive powers is 
■y small. If instead of supposing, as above, that a-=z 
ooooooooi, we suppose it only one millionth part of this 
ue, then the successive powers of i+ff will differ from 
i another by much smaller decimal fractions. 

5. If therefore a be indefinitely small, and successive 
;vers of i +df be raised, a series of numbers in geometrical 
gression will be produced, of which the common numbers 
3, 4, 5, &c. will become terms. For on every multipli* 
ion by I +ff, an indefinitely small addition is made to the 
/ver multiplied, and by this indefinitely, small addition, the 
<t higher power is produced. Some power of i + <z will 
irefore be equal to the number 2, or so nearly equal to it 
.t they may be considered as equali Continuing tjie ad- 

■ icement of the powers of i-\-a, the numbers 3, 4, 5, &c. 
the same reasons, will fall into the series. 

6. The sum of the logarithms of any two numbers is equal 
the logarithm of the product of the same two numbers, 
lus if I + a raised to the «"* power be equal to the number 

^ and \i i-\-a raised to the rrf^ power be equal to the num- 
ber 



1 



494- OF LOGARITHMS. 

ber M, then, by the preceding articles, n is the logarithm 
of I +«^" Or of its equal N, and for the same reason m is the 
logarithm of M. Hence it follows that n-\-m:=zxhc logarithm 

ofNxM, for NxM=7+^''xi"+^''^=iT^"+'^by the 
nature of indices. If the logarithm of N be subtracted 
. from the logarithm of M, the difference is equal to the loga- 
rithm of the quotient which arises from the division of M by 

JN. ror-TT — ==— m+^' , by the nature of indices. — 

vThe addition of logarithms, therefore, answers to the multi- 
plication of the natural numbers to which they belong; and 
the subtraction of logarithms answers to the division by the 
natural numbers to which they belong. 

7. If the logarithms of a series of natural numbers be all mul- 
tiplied by the same number, the several products will have the 
last-mentioned properties of logarithms. Thus if the indices of 
all the powers of i -f o be multiplied by /, then, using the no-j 
tation stated in the last article, the logarithm of N is nl^ and! 
the logarithm of M is ot/, and the logarithm of N x M is «/+ ? 

ml; forNxM=iT^"'xFT^'"'=i'+^"'+''", by the n: 

M M i 
ture of indices. Also ml— nl— the Iogarithm,of -5— , for-r-rrzj 

' ^ „ i — 1~| a^"^^~^- Hence the products arising from the 
i-\-a)" _ - 

multiplication of / into the indices of the powers of i +(?, are 
termed logarithms, as are also all .numbers, which have the 
properties stated at the end of article 6. it is on account of 
these properties that logarithms are so very useful in calcula- 
tions of the highest importance. 

8. If the indices of the powers of'i -f^ be multipled by a, 
the products are called the hyperbolic logarithms of the num- 
bers equal to the powers of i+a. Thus if the number N be 
equal to i"+^"j then na is the hyperbolic logarithm of N 
and if the number M be.equal to i +<3 '", then ffia is the hy- 
perbolic logarithm of M. Hyperbolic logarithms are not those 
in common use, but they can be calculated with less laboui 
than any other kind, and common logarithms are obtainec 
from them. 

9. If successive powers of a very small fraction be raiset 
they will successively be less and less in value. This trutf 



OE LOGARITHMS, 
appears most evident by putt ing the. value in the form of a 

vulgar fraction. Thus — ] = ; \ = 

** IOOOOO| lOOOOOOOOOO lOOOOO) 

. cCC. 

I OOOOOOOOOOOOOOO 

lo. Let it be required to determine the hyperbolic loga- 
rithm L, of any number N. Using the same notation, as in 
the preceding articles, l+tfl"=:N, and, by extracting the n* 

T 

root of each side of the equation, i + a— N". Put ot=:— , 

J_ m n 

and i+A"=N, and then N" zzi+x^ = (by the binomial 

m—i m—i m—2 

theorem) i-\-mx-\-mx X* +»zx x x x^. 

2 2 3 

.-f Sec. =1 +tf. Now as ff is indefinitely small, the power of 
i-f ^, which is equal to the number N, must be indefinitely 
high ; or, which is the same thing, n must be indefinitely 
great. Consequently m must be indefinitely small, and there- 
fore may be rejected from the expressions m — i, w— 2, »i— 3, 
&c. Hence i being taken from each side of the above equa- 

, mx* mx^ mx* mx^ 
tion, we have a—mx— \- 1- — &c. Each side 

^ ^ + 5 ^ ;,* ^t 

of this equation being divided by ot, we have —zzx 1 — 

m 23 

\-~ — &c. But m——, and therefore— =:tf«—.r 1- 

45 n m 2 

Jt' JT* JT* 

— '"7'"" °^^- — ^ ^^ hyperbolic logarithm of N, bjr 

article 8. This series, however, if .r be a whole number, 
does not converge. 

Let M be a whole number and M= , and then x is 

1 ess than I. For multiplying both sides of the equation by 

I— X- we haveM— Mar=:i, and therefore i—rr=r.r. Now 

M 

. , I r 

. t M =: =:i-}-a>;). Then we have i+ar=^=:^i = 

* -* I —x\ 
. — 5- I . > Y \ 

I— .r' ^=: I --.r^'^(by putting rr: ) = !— rr+rX X 

P 2 

r — I r — 2 

*' — r X X X .r' -^ Sec. But for the same reasons 

2 3 

.as 



495 



496 OF LOGARITHMS. 

as above, r must be indefinitely small, and therefore may be 
rejected from the factors r— i, r— 2, r— 3, &c. Conse- 
quently, taking i from each side of the above equa^tion, ^rr 

r.r* rx^ rx'' ^x^ » t> ^ 

— rx — &c. But— r=: — , andthere^r 

_ . 2 3 4 5 P J 

fore dividing the left-hand side of the equation by — , and 

A-i* futS ly/* «5 

the other by — r, we have ep=zx-{- -1— -f ^_ +-1- -{■- {- 

^345 
&c. r: the hyperbolic logarithm of M. 

11. As by the last article, the hyperbolic logarithm of N 

jr* x^ X* x^ x^ x'' 
or i+x is X + 1 ;r- + &c. andasthc 

I X^ X^ X* x^ 

hyperbolic logarithm ofMor — — is .r + -^ — I 1 -| 

I X 2345 

X x'^ I "^* X 

-?--{ V &c. the hyperbolic logarithm of N x M, or 

is equal to the feum of these tw^o series, that is, equal to 2jr + 

2.i'' 2x^ 2.f' 

1 \-— — h &c. This series converges faster than 

either df the preceding, and its value may be expressed thus : 

2X.r + — +— + 1- &c. / 

Z S 1 

7/ -f I 2« + 2 

12. The logarithm of =2 X logarithm of ; 1- 

° n ° 2« + 1 

i — ^* 

the logarithm of ^^=:;-^ — ■• For as the addition of loga- 

2?7+i\ —^ 

rithms ansvv^ers to the multiplication of the numbers to which 
they belong, the logarithm of the square of any number, ts 
the logarithm of the number multiplied by 2. . Hence the lo . 

in 4- 2^1* 2n + 2 2«4-2V 

garithrn of is 2 X logarithm of — ;. But — — — 

2«Tl\ 2'^— I 

2«TT^^ _2« + 2^* __ 4«'-i-8?? + 4 __ «^+2» + I __ 

^2"^l^*-I~2«-f?*-l" 4«' + 4« " «* + « ~ 

n+1 x«+i_ n+i 

« X « + I '^ 

From the preceding articles hyperbqlic logarithms may be 
calculated, as in the following examples. 

Exaropl* 



OF LOGARITHMS. 497 

Example i. Required the hyperbolic logarithm of 2, Put 

»+r , , 2« + 2 4 , 2 n+iV 

~2 and then nm, =— , and .. 

« ' 2«+i 3' 2n+iY-i 

— rr. I order to proceed by the series in article 11, let 
- — —=-", and then .m— . Consequently 

I-T 3 7 

r = 0.14285714286 

— =0.00097181730 

— =0.00001189980 

— =0.00000017347 

— =0.00000000275 
— =0.00000000004 

bum of the above terms, • 0.14384103622 

2 



^ - I + T 2n+2 4 o ,0 

Log. of or or — - 0.28768207244. 

** l—X 2W+I 3 / / *M- 

The double of which is 0.57 5 364 14488, and answers to the 
first part of the expression in article 12. 

secondly, let ~=^ ^^^ then 8 + 8.r =9—9.?', and !s 

I -^ .2' O 



:iow is equal to — . Consequently, 



X =0.05882352941 
x^ ' 

— =0.00006784721 
3 ' 
x^ 

— =0.00000014086 

x'' 

— =0.00000000035 



Sum of the above terms 0.05889151783 



"* J^Tiv^' ""' ^ 0.H778303566, 

^^ which 



4^8 OF LOGARITHMS. 

which answers to the second part of the expression in article 
12. Consequently the hyperbolic logarithm of the number 
2150.57536414488 + 0.1177830356=0.69314718054. 

The hyperbolic logarithm of 2 being thus found, that of 
4, 8, ifc, and all the other powers of 2 may be obtained by 
mutiplying the logarithm of 2 by 2, 3, 4, &c. respectively, 
as is evident from the properties of logarithms stated in ar- 
ticle "6. Thus, br multiplication, the hyperbolic logarithm 
of 4 = 1 . 38629436 1 q8 
of 8=2.07944154162 
&c. ' ■;. 

From the above the logarithm of 3 may easily be obtained. 

For 4-^^=4 X — = 3; and therefore as the logarithm of — 
3 4 3 

was determined above, and also the logarithm of 4. 

From the logarithm of 4, viz. - i. 38629436108, 
Subtract the logarithm of -, viz. 0.28768207244, 



And the logarithm of 3 is - - - - 1.09861228864. 

Having found the logarithms of 2 and 3, we can find, by 
addition only^ the' logarithms of all the powers of 2 and 3, 
and also the logarithms of all the numbers w^hich can be pro- 
duced by multiplication from 2 and 3. Thus, 

To the logarithm of 3, viz. . - - i. 09861228864 

^ Add the logarithrh of 2, viz. - - - 0.693147 18054 

:. -yi:r. ,ii^; ; ^ ', .;.:*( ' r--^^ 

And the sum is the logarithm of 6 - 1.79175946918. 
To this last found add the logarithm of 2, and the sum 
2.48490664972 is the logarithm of 12. 

The hyaerbolic loganthips of other prime numbers may 
be more readily calculated by attending to the following 
article. 

13. Let a, bj c be three numbers in arithmetical progres- 
sion, whose common difference is i. Let b be the prime 
number, whose logarithm is soirght, and a and c even num- 
bers whose logarithms are known, or easily obtained from 
others already computed. Then, a being tke least of the 
three, and the common difference being i, ^7=^—1, and 
c=^+i. Consequently a x c •=.b—ixb + i—b^—ij and 

<r<r+ I =^* J and therefore — = . This is a general ex- 

ac ac 

pression 



^ OF LOGARITHMS. 49f 

prcssion for the fraction, which it will be proper to put = 
—y that the series expressing the hyperbolic logarithm 

may converge quickly. For as = , ac + a:x =. ac 

I —x ac 

+ i—ac.v—x, and therefore 2acx-\-xzzi^ and : = . 

Example 2. Required the hyperbolic logarithm of 5. 

Here a-=.±. rrr6j and .r— ~ — . Consequently, 

iac-\-i 49 ^ -" 

X =0.0204081632 

— =0.0000028332 

— =0.0000000007 

Sum of the above terms, 0.0204109971 

2 



Log of — or— 0.0408219942 

I — -r 24 

2C • ' 

But -^x8x3r=25, and the addition of Logarithms an- 
swers to the multiplication of the natural numbers to which 
they belong. Consequently, 

2^ 

To the log. of -:^ - - - - 0.0408219942 

Add the log. of 8 - - - - -2.0794415422 
And also the log. of 3 - - - - i. 0986122890 

And the sum is the log. of 25 - 3.2188758254 
Thehalf of this, viz. 1.6094379127, is the hyperbolic loga- 
rithm of 5 i for 5 X 5 = 25. 

Example 3. Required the hyperbolic logarithm of 7. 

licre^=r6 <: = ?, and .r= ■— — rr — , and = = 

2 tf c 4- 1 97 as I —X 

— T-. Consequently, 
45 

Kka .r= 



500 ^ OF LOGARITHMS. 

jr =0.01030927835 
—=0.00000036522 

=0.00000000002 

5 

Sum of the terms r - - - .01030964359 

2 



Log. of -^ - - - - - -0.02061928718 

To which add log. of 8 - - - 2.07944154162 
And also log. of 6 - - - - 1.79175946918 

The sum is the log. of 49 - - 3.89182029798 

For ^x6x 8=49. Consequently the half of this, viz. 
48 

1.94591014899, is the hyperbolic logarithm of'7 j for 7 x 7 

If the reader perfectly understand the investigations and 
examples, already given, h'e will find no difficulty in calcu- 
lating the hyperbolic logarithms of kigher prime numbers. 
It will only be necessary for him, in order to guard against 
any embarrassment, to compute them as they advance in 
succession above those already mentioned. Thus, after 
what has been done, it would be proper, first of all, to calcu-. 
late the hyperbolic logarithm of 1 1, then that of 13, &c. 

Proceeding according to the method already explained, it 
will be found that 

The hyperbolic logarithm of II is 2.3978952^3016 

of 1 3 is 2.564999357538 

of 17 IS 2.833213344878 

• of 19 is 2.944438979941 

Logarithms were invented by Lord Neper, Baron of Mer- 
chiston, in Scotland. In the year 1614 he published at Edin- 
burgh a small quarto, containing tables of them, of the hyper- 
bolic kind, and an account of their construction anci use. The 
discovery afforded the highest pleasure to mathematicians, as 
they were fully sensible of the very great utility of logarithms j 
but it was soon suggested by Mr. Briggs, afterwards Saviliari 
Professor of Geometry in Oxford, that another kind of lo- 
garithms would be more convenient, for general purposes, 

than 



OF LOGARITHMS. 501 

than the hyperbolic. That one set of logarithms may be ob- 
tained from another will readily appear from the following 
aiiicle. 

14. It appears from articles i, 3, and 7, that if all the lo- 
garith ms ot the geometrical progression i, i+i?'', T+o}', 
i+fll\ i + a*, I -hfl\% &c. be multiplied or divided by any 
given lumber, the products aod also the quotients will like- 
wise be logarithms, for their addition or subtraction will an- 
swer to the multiplication or division of the terms in the 
geometrical progression to which they belong. The same 
terms in the geometrical progression may therefore be repre- 
sented with different sets or kinds of logarithms in the fol- 
lowinir manner. 



l,i-\-a^\ i+« ) i+a^S i-\-a^, i+a>\ i-{-a\% &c. 
J, rMS i+^H l+«^^i iT^y T+3^7+tfl^ &c. 

I, f+P'^ ! + '»''", i-i-a^'"i TTa^"^ i + fl'"*, T+tfi^j &c. 

In these expressions / and m denote any numbers, whole 
or fractional ; and the positive value of the term in the geo- 
metrical progression, mider the same number in the index, is 
understood to be the same in each of the three series. Thus 

if I +fll* be equal to 7, then i + a ^, is equal to 7, as is also 

4 _ 

i+ai«. If I +a'^ be equal to 10, then i+a'^is equal to 10, 

6 

as is also i +^ *, Sec. If therefore /, 2 /, 3/, &c. be hyperbolic 
logarithms, calculated by the methods already explained, the 

logarithms expressed by — , — ^, -J^, &c. may be derived 

m fn 7n 

from them v for the hyperbolic logarithm of any given num- 
ber is to the logarithm in the last-mentioned set, of the same 

number, in a given ratio. Thus 4/ : — - : : i : -—- zz-j— s 

also 6/ : — : : i : -— -=-r~) ^^* 
m b Im Im 

15. Mr. Briggs's suggestion, above alluded to, was that i 
should be put for the logarithm of 10, and consequently 2 for 
the logarithm of 100, 3 for the logarithm of 1000, &c. This 
proposed alteration appears to have met with the full appro- 
bation of Lord Neper ; and Mr. Briggs afterwards, within- 
credible labour and perseverance, calculated extensive tables 

K k 3 of 



^2 OF LOGARITHMS. 

of logarithms of this new kind, which are naw called com- 
mon logarithms. If the expeditious methods for calculating 
hyperbolic logarithms, explained in the foregoing articles,* 
had been known to Mr. Briggs, his trouble would have been 
comparatively trivial with that which he must have expe- 
rienced in his operations. 

1 6. It has been already determined that the hyperbolic lo- 
garithm of5 is i,6o94379i«7, and that of 2 is 0.69314718054, 
and therefore the sum of these logarithms, viz. 2.30258509324 
is the hyperbolic logarithm of 10, If, therefore, for the sake 
of illustration, as in article 14, we suppose i~+^^— 10, and 

allow, in addition to the hypothesis there formed, that — , 



m 



2 "i A. 

— , -^, — , &c. denote common logarithms, then 6 Izz 
m m m 

2.30258509324, and — — 1 ; and the ratio for reducing the 

hyperbolic logarithm of any number to the common logarithm 
of the same number, is that of 2.30258509324 to i. Thus in 
order to find the common logarithm of 2, 2.39258509324 : 
I : : 0.69314718054': 0.3010299956, the common logarithm 
of 2. The common logarithms of 10 and 2 being known, 
"We obtain the common logarithm of 5, by subtracting the 
common logarithm of 2 from i, the common logarithm of to ; 
for 10 being divided by 2, the quotient is 5. Hence the 
common logarithm of 5 is 0.6989700044. Again, to find 
the common logarithm of 3, 2.30258509324 : i : : 
1.09861228864 : .4771212546, the common logarithm 
*)f3. 

17. As the constant ratio, for the reduction of hyperbolic 
10 common logarithms, is that of 2.30258509324 to i, it is 
evident that the reduction may be made by multiplying the 
hyperbolic logarithm, of the number whose common logarithm 

is sought, by =.4'?42q448i8. 

2.30258509324 

Thus 1. 945910 14899, the hyperbolic logarithm of 7, 
being multiplied by .4342944818, the product, viz. 
.8450980378, &c. is the common logarithm of 7. 

I'he common logarithms of prime numbers being derived 
from the hyperbolic, the common logarithms of other num- 
bers 

• Some of the principal particulars of tlie foregoing method* were discovered by 
the celebrated Ti)om8s Simpsen. Se« als» Mr. Hellins' M»lh«maiical Essays, pub- 
lishedlQ 1788. 



OE LOGARITHMS. 503 

bers may be obtained from those so derived, merely by addi- 
tion or subtraction. For addition of logarithms, in any set 
or kind, answers to the multiplication of the natural num- 
bers to which they belong, and consequent